āĻāϤā§āϤāϰ āĻā§āϞ āĻŽāύ⧠āĻšāϞā§Â āύāĻŋāĻā§ āĻŽāύā§āϤāĻŦā§āϝ āĻāϰā§āύ
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āĻĒāĻĻāĻžāϰā§āĻĨāĻŦāĻŋāĻā§āĻāĻžāύ
Â
Â
ā§§. āϏā§āĻĨāĻŋāϰ āĻ āĻŦāϏā§āĻĨāĻžā§ āĻĨāĻžāĻāĻž āĻāĻāĻāĻŋ āĻŦāϏā§āϤ⧠āĻŦāĻŋāϏā§āĻĢā§āϰāĻŖā§āϰ āĻĢāϞ⧠M1 āĻāĻŦāĻ M2 āĻāϰā§āϰ āĻĻā§āĻāĻŋ āĻāĻŖā§āĻĄā§ āĻŦāĻŋāĻāĻā§āϤ āĻšā§ āĻāĻŦāĻ āĻāĻŖā§āĻĄ āĻĻā§āĻāĻŋ āĻŦāĻŋāĻĒāϰā§āϤ āĻĻāĻŋāĻā§ āϝāĻĨāĻžāĻā§āϰāĻŽā§ V1 āĻāĻŦāĻ V2 āĻŦā§āĻ āĻĒā§āϰāĻžāĻĒā§āϤ āĻšā§āĨ¤ V1 āĻāĻŦāĻ V2 āĻāϰ āĻ āύā§āĻĒāĻžāϤ āĻāϤ?
Â
āĻ. M1 / M2
āĻ. M2 / M1
āĻ. (M1 / M2)1/2
āĻ. (M1 / M2)1/2
Â
āĻāϤā§āϤāϰ : āĻ. M2 / M1
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
                Â
āĻāĻāĻžāύā§, M1V1 = M2V2, â´ V2 / V1 = M2 / M1
Â
⧍. āĻāĻāĻāĻŋ āĻāϰā§āϰ āĻŦāĻŋāĻĒāϰā§āϤ āĻĻā§âāĻĻā§ā§āĻžāϞā§āϰ āĻŽāϧā§āϝāĻŦāϰā§āϤ⧠āĻĻā§āϰāϤā§āĻŦ 4 m, āĻāĻāĻāĻŋ āĻĻā§ā§āĻžāϞ⧠āĻāĻāĻāĻŋ āĻ āĻŦāϤāϞ āĻĻāϰā§āĻĒāĻŖ āϞāĻžāĻāĻžāύā§āĻž āĻāĻā§āĨ¤ āĻĻāϰā§āĻĒāĻŖ āĻšāϤ⧠2.5 m āĻĻā§āϰ⧠āĻāĻāĻāĻŋ āĻŦāϏā§āϤ⧠āϰāĻžāĻāϞ⧠āϤāĻžāϰ āĻĒā§āϰāϤāĻŋāĻŦāĻŋāĻŽā§āĻŦ āĻŦāĻŋāĻĒāϰā§āϤ āĻĻā§ā§āĻžāϞ⧠āĻāĻ āĻŋāϤ āĻšā§āĨ¤ āĻĻāϰā§āĻĒāĻŖā§āϰ āĻĢā§āĻāĻžāϏ āĻĻā§āϰāϤā§āĻŦ āĻāϤ?
Â
āĻ. 2.5 m
āĻ. 1.54 m
āĻ. 1.44 m
āĻ. 2.25 m
Â
āĻāϤā§āϤāϰ : āĻ. 1.54 m
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
 $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}, \therefore \mathrm{f}=\frac{u v}{u+v}=\frac{4 x 2.5}{4+2.5}=1.54$
Â
ā§Š. āĻāĻāĻāĻŋ āϤā§āĻāϏā§āĻā§āϰāĻŋā§ āĻĒāĻĻāĻžāϰā§āĻĨā§āϰ āĻ āϰā§āϧāĻžā§ā§ 1600 āĻŦāĻāϰāĨ¤ āĻāϤ āϏāĻŽā§ āĻĒāϰ⧠āϤā§āĻāϏā§āĻā§āϰāĻŋâā§ āĻĒāĻĻāĻžāϰā§āĻĨā§āϰ 15/16 āĻ āĻāĻļ āĻā§āώā§āĻĒā§āϰāĻžāĻĒā§āϤ āĻšāĻŦā§?
Â
āĻ. 1500 years
āĻ. 4800 years
āĻ. 6400 years
āĻ. 9600 years
Â
āĻāϤā§āϤāϰ : āĻ. 6400 years
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
$\mathrm{t}_{1 / 2}=1600$ āĻŦāĻāϰ,Â
$\therefore \lambda=\frac{0.693}{1600}$
āĻāĻāύ, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ ,
āĻāĻāĻžāύ⧠$\mathrm{N}=(1 / 16) \times \mathrm{N}_{0}, \therefore \mathrm{t}=-\frac{\ln \frac{1}{16}}{(0.693 / 1600)}=6400$ āĻŦāĻāϰ
Â
ā§Ē. Higgs āĻĒā§āϰāĻā§āϰāĻŋā§āĻž āĻāĻ āϧāϰāύā§āϰ -
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āĻ. āĻāϰ āϤā§āϰāĻŋāϰ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž
āĻ. āĻļāĻā§āϤāĻŋ āϤā§āϰāĻŋāϰ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž
āĻ. āĻāϞā§āĻāĻā§āϰāύ āϤā§āϰāĻŋāϰ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž
āĻ. āĻŦāϞ āϤā§āϰāĻŋāϰ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž
Â
āĻāϤā§āϤāϰ : āĻ. āĻāϰ āϤā§āϰāĻŋāϰ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž
Â
āϏā§āϤā§āϰ : Wikipedia
Â
ā§Ģ. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āĻā§āϤ āĻĒā§āϰāĻā§āϰāĻŋā§āĻž āĻļāĻŦā§āĻĻ āϤāϰāĻā§āĻ āĻĻā§āĻŦāĻžāϰāĻž āĻĒā§āϰāĻĻāϰā§āĻļāĻŋāϤ āĻšā§ āύāĻž?
Â
āĻ. āĻĒā§āϰāϤāĻŋāϏāϰāĻŖ
āĻ. āĻŦā§āϝāϤāĻŋāĻāĻžāϰ
āĻ. āϏāĻŽāĻŦāϰā§āϤāύ
āĻ. āĻ āĻĒāĻŦāϰā§āϤāύ
Â
āĻāϤā§āϤāϰ : āĻ. āϏāĻŽāĻŦāϰā§āϤāύ
Â
āϏā§āϤā§āϰ : Wikipedia
Â
ā§Ŧ. āĻĒā§āĻĨāĻŋāĻŦā§āĻĒā§āώā§āĻ ā§ āĻŽāĻšāĻžāĻāϰā§āώā§ā§ āĻĒā§āϰāĻžāĻŦāϞā§āϝ g, āĻāĻžāϞā§āĻĒāύāĻŋāĻ āĻāĻāĻāĻŋ āĻā§āϰāĻšā§āϰ āĻāύāϤā§āĻŦ āϝāĻĻāĻŋ āĻĒā§āĻĨāĻŋāĻŦā§āϰ āĻāύāϤā§āĻŦā§āϰ āϏāĻŽāĻžāύ āĻšā§ āĻāĻŦāĻ āĻŦā§āϝāĻžāϏāĻžāϰā§āϧ āϝāĻĻāĻŋ āĻĻā§āĻŦāĻŋāĻĒā§āĻŖ āĻšā§ āϤāĻŦā§ āĻāĻ āĻā§āϰāĻšā§āϰ āĻĒā§āώā§āĻ ā§ āĻŽāĻšāĻžāĻāϰā§āώā§ā§ āĻā§āώā§āϤā§āϰ āĻĒā§āϰāĻžāĻŦāϞā§āϝ āĻāϤ?
Â
āĻ. g
āĻ. 2g
āĻ. 4g
āĻ. 8g
Â
āĻāϤā§āϤāϰ : 2g
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
$g_{\text {earth }}=\frac{G M}{R^{2}}=\frac{G \rho\left(4 \pi R^{3}\right)}{R^{2}}=G \rho 4 \pi R, \therefore g_{\text {planet }}=G \rho 4 \pi(2 R)=2 \times g_{\text {earth }}$
Â
ā§. āĻāĻāĻāĻŋ āĻŦā§āĻĻā§āϝā§āϤāĻŋāĻ āĻāϏā§āϤā§āϰāĻŋāϤ⧠220 V āĻāĻŦāĻ 1200 W āϞā§āĻāĻž āĻāĻā§āĨ¤ āϝāĻĻāĻŋ āĻĒā§āϰāϤāĻŋ āĻāĻāύāĻŋāĻ āĻŦāĻŋāĻĻā§āϝā§ā§ āĻļāĻā§āϤāĻŋāϰ āĻŽā§āϞā§āϝ 1.00 āĻāĻžāĻāĻž āĻšā§, āϤāĻžāĻšāϞ⧠āĻāϏā§āϤā§āϰāĻŋāĻāĻŋ 2 āĻāĻŖā§āĻāĻž āĻāĻžāϞāĻžāϞ⧠āĻāϤ āĻāϰāĻ āĻĒā§āĻŦā§?
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āĻ. 3 āĻāĻžāĻāĻž
āĻ. 2.6 āĻāĻžāĻāĻž
āĻ. 2.3 āĻāĻžāĻāĻž
āĻ. 2.4 āĻāĻžāĻāĻž
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āĻāϤā§āϤāϰ : āĻ. 2.4 āĻāĻžāĻāĻž
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
P = 1200 W, t = 2 hour, â´ āĻāϰāĻ = (Pt/1000) x 1.00 = 2.4 āĻāĻžāĻāĻž
Â
ā§Ž. āύāĻŋāĻā§āϰ āϞā§āĻāĻāĻŋāϤā§āϰ⧠50 s āϏāĻŽā§āĻāĻžāϞ⧠āĻāĻāĻāĻŋ āĻāĻžā§āĻŋāϰ āĻŦā§āĻā§āϰ āĻĒāϰāĻŋāĻŦāϰā§āϤāύ āĻĻā§āĻāĻžāύ⧠āĻšā§ā§āĻā§āĨ¤ āĻāĻ āϏāĻŽā§āĻāĻžāϞ⧠āĻāĻžā§āĻŋāĻāĻŋ āĻāϤ āĻĻā§āϰāϤā§āĻŦ āĻ āϤāĻŋāĻā§āϰāĻŽ āĻāϰā§āĻā§?
Â
āĻ. 500 m
āĻ. 400 m
āĻ. 350 m
āĻ. 300 m
Â
āĻāϤā§āϤāϰ : āĻ. 350 m
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
āĻĒā§āϰāĻĨāĻŽ ⧍ā§Ļ āϏā§āĻā§āύā§āĻĄā§ ā§§ā§Ļā§Ļ āĻ āϤāĻŋāĻā§āϰāĻŽ āĻāϰā§āĨ¤Â  $\mathrm{S}=\frac{(0+v)}{2} x t=5 \times 20=100$
āĻĒāϰāĻŦāϰā§āϤ⧠⧍ā§Ļ āϏā§āĻā§āύā§āĻĄā§ s = vt = 200
āĻļā§āώ ā§§ā§Ļ āϏā§āĻā§āύā§āĻĄā§ $\mathrm{S}=\frac{(0+v)}{2} x t=5 \times 20=100$, â´ āĻŽā§āĻ = 350 m
Â
⧝. 0°C āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžāϰ 2.1 kg āĻŦāϰāĻĢ 40°C āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžāϰ 5.9 kg āĻĒāĻžāύāĻŋāϰ āϏāĻžāĻĨā§ āĻŽāĻŋāĻļā§āϰāĻŋāϤ āĻāϰāĻž āĻšāϞā§āĨ¤ āĻŽāĻŋāĻļā§āϰāĻŖā§āϰ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻž āĻāϤ āĻšāĻŦā§? āĻĒāĻžāύāĻŋāϰ āĻāĻĒā§āĻā§āώāĻŋāĻ āϤāĻžāĻĒ = 4.2x103 J Kg-1 K-1, āĻŦāϰāĻĢ āĻāϞāύā§āϰ āĻāĻĒā§āĻā§āώāĻŋāĻ āϏā§āĻĒā§āϤāϤāĻžāĻĒ = 3.36x105 J Kg-1 K-1
Â
āĻ. 7.5°C
āĻ. 9.5°C
āĻ. 10.5°C
āĻ. 8.5°C
Â
āĻāϤā§āϤāϰ : āĻ. 8.5°C
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$2.21 \times 3.36 \times 10^{5}+2.21 \times 4.2 \times 10^{3} \times \theta=5.9 \times 4.2 \times 10^{3} \times(40-\theta)$
āϏāĻŽāĻžāϧāĻžāύ āĻāϰā§, θ = 8.5°C
Â
ā§§ā§Ļ. āĻĒāĻžāύāĻŋ āϏāĻžāĻĒā§āĻā§āώ⧠āĻāĻžāĻāĻā§āϰ āĻĒā§āϰāϤāĻŋāϏāĻžāϰāĻā§āĻ 9/8. āĻŦāĻžā§ā§ āϏāĻžāĻĒā§āĻā§āώ⧠āĻāĻžāĻāĻā§āϰ āĻĒā§āϰāϤāĻŋāϏāϰāĻžāĻā§āĻ 3/2, āĻŦāĻžā§ā§ āϏāĻžāĻĒā§āĻā§āώ⧠āĻĒāĻžāύāĻŋāϰ āĻĒā§āϰāϤāĻŋāϏāĻžāϰāĻā§āĻ āĻāϤ?
Â
āĻ. 2/3
āĻ. 4/5
āĻ. 4/3
āĻ. 3/4
Â
āĻāϤā§āϤāϰ : āĻ. 4/3
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$w \mu_{g}=a \mu_{g} / a \mu_{W}, \therefore a \mu_{W}=a \mu_{g} / w \mu_{g}=\frac{\frac{3}{2}}{\frac{9}{8}}=\frac{3 x 8}{2 x 9}=\frac{4}{3}$
Â
ā§§ā§§. āϏāϰāϞ āĻāύā§āĻĻāĻŋāϤ āĻāϤāĻŋāϤ⧠āĻāϞāĻŽāĻžāύ āĻāĻāĻāĻŋ āĻŦāϏā§āϤā§āϰ āĻŽā§āĻ āĻļāĻā§āϤāĻŋ E, āĻāĻŽā§āĻĒāĻžāĻā§āĻ āĻ āĻĒāϰāĻŋāĻŦāϰā§āϤāĻŋāϤ āϰā§āĻā§ āĻŦāĻŋāϏā§āϤāĻžāϰ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻāϰāϞ⧠āϏāϰāϞ āĻāύā§āĻĻāĻŋāϤ āĻāϤāĻŋāϤ⧠āĻāϞāĻŽāĻžāύ āĻŦāϏā§āϤā§āĻāĻŋāϰ āĻŽā§āĻ āĻāϤāĻŋāĻļāĻā§āϤāĻŋ āĻāϤ āĻšāĻŦā§?
Â
āĻ. E
āĻ. 2E
āĻ. E/2
āĻ. 4E
Â
āĻāϤā§āϤāϰ : āĻ. 4E
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
āĻāϤāĻŋ āĻļāĻā§āϤāĻŋ, k = ÂŊ K (A2 â x2), āĻ āϰā§āĻĨāĻžā§ āĻāϤāĻŋāĻļāĻā§āϤāĻŋ āĻŦāĻŋāϏā§āϤāĻžāϰā§āϰ āĻŦāϰā§āĻā§āϰ āϏāĻŽāĻžāύā§āĻĒāĻžāϤāĻŋāĻ
â´ A āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻāϰāĻž āĻšāϞ⧠āĻŽā§āĻ āĻāϤāĻŋāĻļāĻā§āϤāĻŋ (22)xE= 4E
Â
⧧⧍. āĻšāĻžāĻāĻĄā§āϰā§āĻā§āύ āĻĒāϰāĻŽāĻžāĻŖā§āϰ āĻļāĻā§āϤāĻŋ āϏā§āϤāϰā§āϰ āĻĒā§āϰāĻāĻžāĻļ $\mathrm{E}=-\frac{13.5}{n^{2}} \mathrm{eV},(\mathrm{n}=1.2, \ldots)$. āĻā§āĻŽāĻŋ āĻ āĻŦāϏā§āĻĨāĻž āĻĨā§āĻā§ āĻĒāϰāĻŦāϰā§āϤ⧠āĻāĻā§āĻāϤāϰ āĻļāĻā§āϤāĻŋāϏā§āϤāϰ⧠āϝā§āϤ⧠āĻāĻāĻāĻŋ āĻšāĻžāĻāĻĄā§āϰā§āĻā§āύ āĻĒāϰāĻŽāĻžāĻŖā§ āĻāĻŋ āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻļāĻā§āϤāĻŋ āĻļā§āώāĻŖ āĻāϰā§?
Â
āĻ. 3.4 eV
āĻ. 4.5 eV
āĻ. 10.2 eV
āĻ. 13.6 eV
Â
āĻāϤā§āϤāϰ : āĻ. 3.4 eV
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
n = 2 āĻšāϞā§Â $\mathrm{E}=\frac{13.5}{2^{2}} \mathrm{eV}=3 \cdot 4 \mathrm{eV}$
Â
ā§§ā§Š. āĻāĻāĻāĻŋ āĻāĻŖāĻžāϰ āĻāϰāĻŦā§āĻ P, āĻāĻŖāĻžāĻāĻŋāϰ āĻāϤāĻŋāĻļāĻā§āϤāĻŋ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻāϰāĻž āĻšāϞ⧠āĻāϰ āύāϤā§āύ āĻāϰāĻŦā§āĻ āĻāϤ āĻšāĻŦā§?
Â
āĻ. 2P
āĻ. 2P
āĻ. 4P
āĻ. 8P
Â
āĻāϤā§āϤāϰ : āĻ. 2P
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{E}=\frac{1}{2} \mathrm{x} \frac{p^{2}}{m}, \mathrm{E} \propto \mathrm{P}^{2} \therefore 2 \mathrm{E}$ āĻāϰ āĻāύā§āϝ āύāϤā§āύ āĻāϰāĻŦā§āĻ (â2P)
Â
ā§§ā§Ē. 100 W āĻā§āώāĻŽāϤāĻž āϏāĻŽā§āĻĒāύā§āύ āĻāĻāĻāĻŋ āĻšāĻŋāĻāĻžāϰ⧠2 kg āĻāϰā§āϰ āĻāĻāĻāĻŋ āĻāĻĒāĻžāϰā§āϰ āĻāĻŖā§āĻĄāĻā§ 40 s āϤāĻžāĻĒ āĻĻā§ā§āĻž āĻšāϞ⧠āĻāĻŖā§āĻĄāĻāĻŋāϰ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻž āĻāϤ āĻŦā§āĻĻā§āϧāĻŋ āĻšāĻŦā§? āĻāĻĒāĻžāϰā§āϰ āĻāĻĒā§āĻā§āώāĻŋāĻ āϤāĻžāĻĒ 400 J/(Kg K)
Â
āĻ. 5 K
āĻ. 10 K
āĻ. 20 K
āĻ. 50 K
Â
āĻāϤā§āϤāϰ : āĻ. 5 K
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
$\mathrm{Q}=\mathrm{Pt}=100 \times 40=4000, \therefore \Delta \theta=\frac{4000}{2 \times 400}=5 \mathrm{~K}$
Â
ā§§ā§Ģ. 300 Hz āĻāĻŽā§āĻĒāĻžāĻā§āĻā§āϰ āĻāĻŦāĻ āĻŦāĻŋāĻĒāϰā§āϤ āĻĻāĻŋāĻā§ āĻ āĻā§āϰāĻāĻžāĻŽā§ āĻĻā§āĻāĻŋ āĻ āĻāĻŋāύā§āύ āϤāϰāĻā§āĻā§āϰ āĻāĻĒāϰāĻŋāĻĒāĻžāϤāύā§āϰ āĻĢāϞ⧠āĻāĻāĻāĻŋ āϏā§āĻĨāĻŋāϰ āϤāϰāĻā§āĻā§āϰ āϏā§āώā§āĻāĻŋ āĻšā§ā§āĻā§āĨ¤ āϏā§āĻĨāĻŋāϰ āϤāϰāĻā§āĻā§āϰ āĻĒāϰ āĻĒāϰ āĻĻā§āĻāĻŋ āύāĻŋāϏā§āĻĒāύā§āĻĻ āĻŦāĻŋāύā§āĻĻā§āϰ āĻĻā§āϰāϤā§āĻŦ 1.5 m. āĻ āĻā§āϰāĻāĻžāĻŽā§ āϤāϰāĻā§āĻ āĻĻā§āĻāĻŋāϰ āĻŦā§āĻ āĻāϤ?
Â
āĻ. 100 m/s
āĻ. 200 m/s
āĻ. 450 m/s
āĻ. 900 m/s
Â
āĻāϤā§āϤāϰ : āĻ. 900 m/s
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
        Â
Â
ā§§ā§Ŧ. āĻāĻāĻāĻŋ āĻā§āĻļāĻŋāĻ āύāϞā§āϰ āĻŦā§āϝāĻžāϏ 0.04x10-4m. āĻāϰ āĻāĻ āĻĒā§āϰāĻžāύā§āϤ āĻĒāĻžāύāĻŋāϤ⧠āĻĄā§āĻŦāĻžāϞ⧠āĻĒāĻžāύāĻŋ āύāϞā§āϰ āĻāĻŋāϤāϰ 0.082 m āĻāĻĒāϰ⧠āĻāĻ ā§āĨ¤ āĻĒāĻžāύāĻŋāϰ āϤāϞ āĻāĻžāύ āĻāϤ? āĻĻā§āĻā§āĻž āĻāĻā§, āϏā§āĻĒāϰā§āĻļ āĻā§āĻŖ = 0°C āĻāĻŦāĻ āĻĒāĻžāύāĻŋāϰ āĻāύāϤā§āĻŦ = 1.0x103 Kg/m3
Â
āĻ. 8.5 x 10-4 N/m
āĻ. 7.5 x 10-4 N/m
āĻ. 9.0 x 10-4 N/m
āĻ. 8.0 x 10-4 N/m
Â
āĻāϤā§āϤāϰ : āĻ. 8.0 x 10-4 N/m
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{T}=\frac{\mathrm{hr} \rho \mathrm{g}}{2 \cos \theta}, \mathrm{T}=\frac{0.082 \times 0.02 \times 10^{-4} \times 10^{3} \times 9.8}{2}=8.0 \times 10^{-4}$
Â
Â
ā§§ā§. āĻāĻāĻāĻŋ āϧāĻžāϰāĻā§āϰ āĻĻā§āĻ āĻĒāĻžāϤā§āϰ āĻŽāϧā§āϝ⧠āĻŦāĻŋāĻāĻŦ āĻĒāĻžāϰā§āĻĨāĻā§āϝ V āĻāĻŦāĻ āϧāĻžāϰāĻā§āϰ āϏāĻā§āĻāĻŋāϤ āĻļāĻā§āϤāĻŋ X, āϧāĻžāϰāĻā§āϰ āĻŦāĻŋāĻāĻŦ āĻĒāĻžāϰā§āĻĨāĻā§āϝ āĻŦā§āĻĻā§āϧāĻŋ āĻāϰ⧠3V āĻāϰāĻž āĻšāϞ⧠āϏāĻā§āĻāĻŋāϤ āĻļāĻā§āϤāĻŋ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§ā§ā§ āĻāϤ āĻšāĻŦā§?
Â
āĻ. 3 X
āĻ. 6 X
āĻ. 9 X
āĻ. 27 X
Â
āĻāϤā§āϤāϰ : āĻ. 9 X
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}, 3 \mathrm{~V}$ āĻāϰāĻž āĻšāϞ⧠āϏāĻā§āĻāĻŋāϤ āĻļāĻā§āϤāĻŋ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§ā§ā§ āĻšāĻŦā§ $9X$
Â
Â
ā§§ā§Ž. āĻāĻāĻāĻŋ ${ }^{238}{ }_{93} \mathrm{~U}$  āύāĻŋāĻāĻā§āϞāĻŋā§āĻžāϏ āĻĻā§āĻ āϧāĻžāĻĒā§ āĻā§āώ⧠āĻšā§ā§ ${ }^{234}{ }_{91} \mathrm{~Pa}$ āύāĻŋāĻāĻā§āϞāĻŋā§āĻžāϏ āϏā§āώā§āĻāĻŋ āĻāϰā§āĨ¤ āĻāĻ āĻĻā§āĻ āϧāĻžāĻĒā§ āĻā§ āĻā§ āϧāϰāύā§āϰ āϰāĻļā§āĻŽāĻŋ āύāĻŋāϰā§āĻāϤ āĻšā§?
Â
āĻ. Îą and Ã
āĻ. Îą and Îŗ
āĻ. à and Ã
āĻ. à and Îŗ
Â
āĻāϤā§āϤāϰ :
Â
⧧⧝. āĻāĻāĻāĻŋ āĻāĻŽāύ āĻāĻŽāĻŋāĻāĻžāϰ āĻā§āϰāĻžāύāĻāĻŋāϏā§āĻāĻžāϰā§āϰ à = 100 āĻāĻŦāĻ IB = 50 ÂĩA āĻšāĻ˛ā§ Îą āĻāϤ?
Â
āĻ. 1.01
āĻ. 0.99
āĻ. 1.00
āĻ. 1.10
Â
āĻāϤā§āϤāϰ : āĻ. 0.99
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
$\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \therefore \mathrm{I}_{\mathrm{C}}=100 \times 50 \times 10^{-6}=0.005, \therefore \mathrm{I}_{\mathrm{E}}=0.00005+0.005=0.00505$,
$\therefore \alpha=\frac{0.005}{0.00505}=0.99$
Â
⧍ā§Ļ. āĻāĻāĻāĻŋ āĻāĻžā§āĻŋ āϏā§āĻāĻž āĻāϤā§āϤāϰ āĻĻāĻŋāĻā§ 90 m āĻĒāĻĨ 15 s āϏāĻŽā§ā§ āĻ āϤāĻŋāĻā§āϰāĻŽ āĻāϰā§āĨ¤ āĻĒāϰāĻŦāϰā§āϤā§āϤ⧠āĻāĻžā§āĻŋāĻāĻŋ āĻĻā§āϰā§āϤ āĻā§āϰ⧠āĻĻāĻā§āώāĻŋāĻŖ āĻĻāĻŋāĻā§ 40 m āĻĻā§āϰāϤā§āĻŦ 5 s āϏāĻŽā§ā§ āĻ āϤāĻŋāĻā§āϰāĻŽ āĻāϰā§āĨ¤ āĻāĻ 20 s āϏāĻŽā§āĻāĻžāϞ⧠āĻāĻžā§āĻŋāĻāĻŋāϰ āĻā§ āĻŦā§āĻā§āϰ āĻŽāĻžāύ āĻāϤ?
Â
āĻ. 2.5 m/s
āĻ. 5.0 m/s
āĻ. 6.5 m/s
āĻ. 7.0 m/s
Â
āĻāϤā§āϤāϰ : āĻ. 6.5 m/s
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
130/20 = 6.5
Â
⧍⧧. āĻāĻāĻāĻŋ āϤāĻžāϰā§āϰ āĻāĻĒāϰ āĻāĻžāύ F āĻšāϞ⧠āĻĻā§āϰā§āĻā§āϝāĻŦā§āĻĻā§āĻāĻŋ āĻšā§ x, āϤāĻžāϰāĻāĻŋ āϝāĻĻāĻŋ āĻšā§āĻā§āϰ āϏā§āϤā§āϰ āĻŽā§āύ⧠āĻāϞ⧠āĻāĻŦāĻ āϤāĻžāϰā§āϰ āĻāĻĒāĻžāĻĻāĻžāύā§āϰ āĻā§āĻ āĻā§āĻŖāĻžāĻā§āĻ Y āĻšā§ āϤāĻŦā§ āϤāĻžāϰ⧠āϏāĻā§āĻāĻŋāϤ āĻŦāĻŋāĻāĻŦ āĻļāĻā§āϤāĻŋ āĻāϤ?
Â
A. $\frac{1}{2} \mathrm{Yx}$
B. $\mathrm{Yx}$
C. $\frac{1}{2} \mathrm{Fx}$
D. $\mathrm{Fx}$
Â
āĻāϤā§āϤāϰ :  C. $\frac{1}{2} \mathrm{Fx}$
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
                 $w=\frac{1}{2} \times \frac{Y A l^{2}}{L}$ , āĻāĻāĻžāύā§, $\mathrm{Y}=\frac{F L}{A x}$  āĻŦāĻž
$\frac{A}{\mathrm{~L}}=\frac{F}{Y x}, \mathrm{~W}=\frac{1}{2} \mathrm{x} \frac{Y F x^{2}}{Y x}=\frac{1}{2} \mathrm{Fx}$
Â
⧍⧍. āϤā§āĻŽāĻžāϰ āĻāĻāĻāĻŋ 15Ί āϰā§āϧ āĻĒā§āϰā§ā§āĻāύ āĻāĻŋāύā§āϤ⧠āϤā§āĻŽāĻžāϰ āĻāĻžāĻā§ āĻā§ā§āĻāĻāĻŋ 10Ί āϰā§āϧ āĻāĻā§āĨ¤ āĻā§āĻāĻžāĻŦā§ āϤā§āĻŽāĻŋ 10Ί āϰā§āϧ āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻāϰ⧠15Ί āϰā§āϧāĻāĻŋ āϤā§āϰāĻŋ āĻāϰāĻŦā§?
Â
āĻ. āϤāĻŋāύāĻāĻŋ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āϏāĻāϝā§āĻā§
āĻ. āϤāĻŋāύāĻāĻŋ āĻļā§āϰā§āĻŖā§āĻŦāĻĻā§āϧ āϏāĻāϝā§āĻā§
āĻ. āĻĻā§āĻāĻŋ āĻļā§āϰā§āĻŖā§āĻŦāĻĻā§āϧ āϏāĻāϝā§āĻā§
āĻ. āĻĻā§āĻāĻŋ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āϏāĻāϝā§āĻā§ āĻāĻŦāĻ āĻāĻāĻāĻŋ āĻļā§āϰā§āĻŖāĻŋāĻŦāĻĻā§āϧ āϏāĻāϝā§āĻā§
Â
āĻāϤā§āϤāϰ : āĻ. āĻĻā§āĻāĻŋ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āϏāĻāϝā§āĻā§ āĻāĻŦāĻ āĻāĻāĻāĻŋ āĻļā§āϰā§āĻŖāĻŋāĻŦāĻĻā§āϧ āϏāĻāϝā§āĻā§
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$10+\frac{1}{10^{-1}+10^{-1}}=15$
Â
ā§¨ā§Š. āĻŦāϞāĻŦāĻŋāĻĻā§āϝāĻžāϰ āĻŦāĻŋāĻāĻŋāύā§āύ āĻŽā§āϞāĻŋāĻ āĻā§āϤ āϰāĻžāĻļāĻŋ āϏāĻŽā§āĻš āĻšāϞ?
Â
āĻ. āĻāϰ, āĻŦāϞ āĻāĻŦāĻ āϏāĻŽā§
āĻ. āĻāϰ, āĻĻā§āϰā§āĻā§āϝ āĻāĻŦāĻ āϏāĻŽā§
āĻ. āĻŦāϞ, āĻļāĻā§āϤāĻŋ āĻāĻŦāĻ āϏāĻŽā§
āĻ. āĻŦāϞ, āĻāϰ āĻāĻŦāĻ āϏāĻŽā§
Â
āĻāϤā§āϤāϰ : āĻ. āĻāϰ, āĻĻā§āϰā§āĻā§āϝ āĻāĻŦāĻ āϏāĻŽā§
Â
⧍ā§Ē. āĻāĻāĻ āĻĻā§āϰā§āĻā§āϝ āĻāĻŦāĻ āĻāĻāĻ āĻĒāĻĻāĻžāϰā§āĻĨ āĻĻāĻŋā§ā§ āϤā§āϰāĻŋ āĻĻā§āĻāĻŋ āϤāĻžāϰ P āĻāĻŦāĻ Q āĻā§ āĻāĻāĻāĻŋ āĻŦā§āϝāĻžāĻāĻžāϰāĻŋāϰ āϏāĻžāĻĨā§ āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ⧠āϏāĻāϝā§āĻā§āϤ āĻāϰāĻž āĻšā§ā§āĻā§āĨ¤ P āϤāĻžāϰā§āϰ āĻŦā§āϝāĻžāϏ 2mm āĻāĻŦāĻ Q āϤāĻžāϰā§āϰ āĻŦā§āϝāĻžāϏ 1mm. P āĻāĻŦāĻ Q āĻāϰ āϤā§āĻŋā§ āĻĒā§āϰāĻŦāĻžāĻšā§āϰ āĻ āύā§āĻĒāĻžāϤ āĻāϤ?
Â
āĻ. 1/4
āĻ. 1/2
āĻ. 2/1
āĻ. 4/1
Â
āĻāϤā§āϤāϰ : āĻ. 4/1
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\frac{l_{1}}{l_{2}}=\frac{R_{2}}{R_{1}}=\frac{A_{1}}{A_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{4}{1}$
Â
⧍ā§Ģ. 5 kg āĻāϰā§āϰ āĻāĻāĻāĻŋ āϰāĻžāĻāĻĢā§āϞ āĻĨā§āĻā§ 20 g āĻāϰā§āϰ āĻāĻāĻāĻŋ āĻŦā§āϞā§āĻ 1000 m/s āĻāϤāĻŋāϤ⧠āĻā§āĻā§ āϝāĻžā§āĨ¤ āĻĒāĻŋāĻāύ āĻĻāĻŋāĻā§ āϰāĻžāĻāĻĢā§āϞā§āϰ āϧāĻžāĻā§āĻāĻžāϰ āĻŦā§āĻ āĻāϤ?
Â
āĻ. 4 m/s
āĻ. 4000 m/s
āĻ. 400 m/s
āĻ. 40 m/s
Â
āĻāϤā§āϤāϰ : āĻ. 4 m/s
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{V}=\frac{0.02 \times 1000}{5}=4$
Â
⧍ā§Ŧ. āĻāĻāĻāĻŋ āĻĒāĻžāĻĨāϰāĻā§ āϏā§āĻĨāĻŋāϰ āĻ āĻŦāϏā§āĻĨāĻžā§ āĻāĻāĻāĻŋ āĻāĻāĻā§ āĻĻāĻžāϞāĻžāύ āĻĨā§āĻā§ āĻā§ā§ā§ āĻĻā§āĻā§āĻž āĻšāϞāĨ¤ āĻā§āĻŽāĻŋāϤ⧠āĻĒā§āĻāĻžāϤ⧠āĻĒāĻžāĻĨāϰāĻāĻŋāϰ 4 s āĻāϰ āĻŦā§āĻļāĻŋ āϏāĻŽā§ āϞāĻžāĻā§āĨ¤ āĻŦāĻžāϤāĻžāϏā§āϰ āĻāϰā§āώāĻŖ āĻā§āώā§āĻĻā§āϰ āĻšāϞ⧠āĻĒāĻžāĻĨāϰāĻāĻŋāϰ āĻĒā§āϰāĻĨāĻŽ 4 s āϏāĻŽā§ā§ āĻĒāϤāύā§āϰ āĻĻā§āϰāϤā§āĻŦ āĻāĻŦāĻ āĻĒā§āϰāĻĨāĻŽ 2 s āϏāĻŽā§ā§ āĻĒāϤāύā§āϰ āĻĻā§āϰāϤā§āĻŦā§āϰ āĻ āύā§āĻĒāĻžāϤ āĻāϤ?
Â
āĻ. 1/4
āĻ. 4/1
āĻ. 1/2
āĻ. 2/1
Â
āĻāϤā§āϤāϰ : āĻ. 4/1
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}, \therefore \frac{h_{1}}{h_{2}}=\left(\frac{t_{1}}{t_{2}}\right)^{2}=\frac{4}{1}$
Â
⧍ā§. āĻāĻŋāϤā§āϰ⧠āĻāĻāĻāĻŋ āĻŦāϰā§āϤāύā§āϤ⧠āϏāĻŽāĻžāύā§āϤāϰāĻžāϞ āϏāύā§āύāĻŋāĻŦā§āĻļā§ āϏāĻāϝā§āĻā§āϤ āϤāĻŋāύāĻāĻŋ āϰā§āϧ āĻĻā§āĻāĻžāύ⧠āĻšā§ā§āĻā§āĨ¤ āĻŦā§āϝāĻžāĻāĻžāϰāĻŋāϰ āϤā§āĻŋā§-āĻāĻžāϞāĻ āĻļāĻā§āϤāĻŋ 12 V āĻāĻŦāĻ āĻ āĻā§āϝāύā§āϤāϰā§āĻŖ āϰā§āϧ āύāĻāύā§āϝāĨ¤ āĻ ā§āϝāĻžāĻŽāĻŋāĻāĻžāϰā§āϰ āĻĒāĻžāĻ 3.2 A āĻšāϞā§, X āĻāϰ āϰā§āϧ āĻāϤ?
Â
Â
āĻ. 2.1 Ί
āĻ. 4.6 Ί
āĻ. 6.0 Ί
āĻ. 15 Ί
Â
āĻāϤā§āϤāϰ : āĻ. 15 Ί
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{R}=\frac{12}{3.2}=3.75$ , āĻāĻŦāĻžāĻ°Â $\frac{1}{10^{-1}+10^{-1}+15^{-1}}=3.75$
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ā§¨ā§Ž. āĻāĻāĻāĻŋ āĻŦāϏā§āϤā§āĻā§ āĻ āύā§āĻā§āĻŽāĻŋāϰ āϏāĻžāĻĨā§ 30° āĻā§āĻŖā§ āύāĻŋāĻā§āώā§āĻĒ āĻāϰāĻž āĻšāϞāĨ¤ āĻĒāϰāĻŦāϰā§āϤā§āϤ⧠āĻāĻāĻ āĻŦāϏā§āϤā§āĻā§ āĻāĻāĻ āĻāĻĻāĻŋ āĻĻā§āϰā§āϤāĻŋāϤ⧠āĻ āύā§āĻā§āĻŽāĻŋāϰ āϏāĻžāĻĨā§ 40° āĻā§āĻŖā§ āύāĻŋāĻā§āώā§āĻĒ āĻāϰāĻž āĻšāϞāĨ¤ āύāĻŋāĻŽā§āύā§āϰ āĻā§āύāĻāĻŋ āϏāϤā§āϝ āύā§
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āĻ. āĻ āύā§āĻā§āĻŽāĻŋāĻ āĻĒāĻžāϞā§āϞāĻž āĻŦā§āĻĻā§āĻāĻŋ āĻĒā§āϞ
āĻ. āĻŦā§āĻā§āϰ āĻ āύā§āĻā§āĻŽāĻŋāĻ āĻāĻĒāĻžāĻāĻļ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§āϞ
āĻ. āϏāϰā§āĻŦā§āĻā§āĻ āĻāĻā§āĻāϤāĻž āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§āϞ
āĻ. āĻŦāϏā§āϤā§āĻāĻŋāϰ āĻāĻĄā§āĻĄā§āύāĻāĻžāϞ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§āϞ
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āĻāϤā§āϤāϰ : āĻ. āĻŦā§āĻā§āϰ āĻ āύā§āĻā§āĻŽāĻŋāĻ āĻāĻĒāĻžāĻāĻļ āĻŦā§āĻĻā§āϧāĻŋ āĻĒā§āϞ
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
āĻŦā§āĻā§āϰ āĻ āύā§āĻā§āĻŽāĻŋāĻ āĻāĻĒāĻžāĻāĻļ, $\mathrm{V}_{0} \cos \theta_{0}$ , āĻāĻŦāĻ $\cos 30^{\circ}>\cos 40^{\circ}$
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⧍⧝. āĻāĻāĻāĻŋ āĻāĻĻāϰā§āĻļ āĻā§āϰāĻžāύā§āϏāĻĢāϰā§āĻŽāĻžāϰā§āϰ āĻā§āĻŖ āĻ āĻŽā§āĻā§āϝ āĻā§āĻŖā§āĻĄāϞā§āϰ āĻĒāĻžāĻ āϏāĻāĻā§āϝāĻžāϰ āĻ āύā§āĻĒāĻžāϤ 6:1. āϝāĻĻāĻŋ āĻŽā§āĻā§āϝ āĻā§āĻŖā§āĻĄāϞā§āϤ⧠āĻĒā§āϰāϤāĻŋ āϏā§āĻā§āύā§āĻĄā§ āĻŦā§āϝā§āĻŋāϤ āĻļāĻā§āϤāĻŋ 6 J āĻšā§, āϤāĻŦā§ āĻā§āĻŖ āĻā§āĻŖā§āĻĄāϞā§āϤ⧠āĻŦā§āĻĻā§āϝā§āϤāĻŋāĻ āĻā§āώāĻŽāϤāĻž āĻāϤ?
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āĻ. 6 J
āĻ. 36 Js-1
āĻ. 6 W
āĻ. 36 W
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āĻāϤā§āϤāϰ : āĻ. 6 W
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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āĻŽā§āĻā§āϝ āĻ āĻā§āĻŖ āĻā§āĻŖā§āĻĄāϞā§āϤ⧠āĻŦā§āĻĻā§āϝā§āϤāĻŋāĻ āĻā§āώāĻŽāϤāĻž āϏāĻŽāĻžāύāĨ¤
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ā§Šā§Ļ. m āĻāϰā§āϰ āĻāĻāĻāĻŋ āĻŦāϏā§āϤ⧠r āĻŦā§āϝāĻžāϏāĻžāϰā§āϧā§āϰ āĻŦā§āϤā§āϤāĻžāĻāĻžāϰ āĻĒāĻĨā§ āϏāĻŽāĻĻā§āϰā§āϤāĻŋāϤ⧠āĻāϞāĻā§āĨ¤ āĻŦā§āϤāĻžāĻāĻžāϰ āĻāϤāĻŋāϰ āĻĒāϰā§āϝāĻžā§āĻāĻžāϞ T, āĻŦāϏā§āϤā§āĻāĻŋāϰ āĻāĻĒāϰ āĻā§āύā§āĻĻā§āϰāĻŽā§āĻā§ āĻŦāϞā§āϰ āĻŽāĻžāύ āĻāϤ?
āĻ. 4Ī2mr/T2
āĻ. 4Ī2mr/T
āĻ. 4Īmr2/T2
āĻ. Īmr2
āĻāϤā§āϤāϰ : āĻ. 4Ī2mr/T2
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{F}=\mathrm{m} \alpha, \alpha=\omega^{2} \mathrm{r}=\left(\frac{2 \pi}{\tau}\right)^{2} \times r \therefore F=\frac{4 \pi^{2} m r}{\tau^{2}}$
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āĻāĻŖāĻŋāϤ
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1. y = mx, y = m1x āĻāĻŦāĻ y = b āϏāϰāϞāϰā§āĻāĻžāϤā§āϰ⧠āĻĻā§āĻŦāĻžāϰāĻž āĻāĻ āĻŋāϤ āϤā§āϰāĻŋāĻā§āĻā§āϰ āĻŦāϰā§āĻāĻāĻāĻā§ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻšāĻŦā§
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(A) $\frac{\mathrm{b}^{2}\left(\mathrm{~m}_{1}-\mathrm{m}\right)}{2 \mathrm{~mm}_{1}}$
(B) $\frac{\mathrm{b}^{2}\left(\mathrm{~m}-\mathrm{m}_{1}\right)}{2 \mathrm{~mm}_{1}}$
(C) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{\mathrm{mm}_{1}}$
(D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$
Answer: (D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :Â
$\mathrm{O} \equiv(\mathrm{o}, \mathrm{o})$
$\mathrm{A} \equiv\left(\frac{\mathrm{b}}{\mathrm{m}}, \mathrm{b}\right)$
$\mathrm{B} \equiv\left(\frac{\mathrm{b}}{\mathrm{m}_{1}}, \mathrm{~b}\right)$
$\Delta \mathrm{OAB}=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \frac{\mathrm{b}}{\mathrm{m}} & \mathrm{b} & 1 \\ \frac{\mathrm{b}}{\mathrm{m}_{1}} & \mathrm{~b} & 1\end{array}\right|=\frac{1}{2}\left(\frac{\mathrm{b}^{2}}{\mathrm{~m}}-\frac{\mathrm{b}^{2}}{\mathrm{~m}_{1}}\right)=\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$
Answer: (D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$
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2. 3x2 + 5y2 = 15 āĻāĻĒāĻŦā§āϤā§āϤā§āϰ āĻā§āĻā§āύā§āĻĻā§āϰāĻŋāĻāϤāĻž āĻšāĻŦā§
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(A) $\sqrt{3 / 5}$
(B) $\sqrt{5 / 3}$
(C) $\sqrt{2 / 5}$
(D) $\sqrt{5 / 2}$
Answer: (C) $\sqrt{2 / 5}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :Â
$3 x^{2}+5 y^{2}=15$
$\Rightarrow \frac{3 x^{2}}{15}+\frac{5 y^{2}}{15}=0$
$\Rightarrow \frac{x^{2}}{5}+\frac{y^{2}}{3}=0$
$\therefore a^{2}=5, b^{2}=3$
$\because a>b$
$\mathrm{e}^{2}=\frac{\mathrm{a}^{2}-\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{5-3}{5}=\frac{2}{5}$
$\therefore e=\sqrt{\frac{2}{5}}$
Answer: (C) $\sqrt{2 / 5}$
3. $\left(2 x^{2}-\frac{1}{2 x^{3}}\right)^{10}$ āĻāϰ āĻŦāĻŋāϏā§āϤā§āϤāĻŋāϤ⧠x-āĻŦāϰā§āĻāĻŋāϤ āĻĒāĻĻāĻāĻŋ āĻāϤāϤāĻŽ āĻāĻŦāĻ āĻāϰ āĻŽāĻžāύ āĻāϤ?
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(A) āĻĒāĻā§āĻāĻŽ āĻāĻŦāĻ 840
(B) āĻāϤā§āϰā§āĻĨ āĻāĻŦāĻ 1920
(C) āώāώā§āĻ āĻāĻŦāĻ 252
(D) āϏāĻĒā§āϤāĻŽ āĻāĻŦāĻ 30
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Answer: (A) āĻĒāĻā§āĻāĻŽ āĻāĻŦāĻ 840
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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(r + 1) āϤāĻŽ āĻĒāĻĻ āĻšāĻŦā§
${ }^{10} \mathrm{C}_{\mathrm{r}}\left(2 \mathrm{x}^{2}\right)^{10-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}^{3}}\right)^{\mathrm{r}}$
$=10 \mathrm{C}_{\mathrm{r}} \cdot 2^{10-\mathrm{r}} \cdot \mathrm{x}^{20-2 \mathrm{r}} \cdot(-1)^{\mathrm{r}} \cdot 2^{-\mathrm{r}} \cdot \mathrm{x}^{-3 \mathrm{r}}$
$=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}} \cdot 2^{10-2 \mathrm{r}} \cdot \mathrm{x}^{20-5 \mathrm{r}}$
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āϝāĻĻāĻŋ āĻĒāĻĻāĻāĻŋ x āĻŦāϰā§āĻāĻŋāϤ āĻšā§ āϤāĻŦā§
$20-5 r=0 \Rightarrow 5 r=20 \Rightarrow r=4$
â´ (r + 1) = 5-āϤāĻŽ āĻĒāĻĻ x āĻŦāϰā§āĻāĻŋāϤ āĻāĻŦāĻ āĻāϰ āĻŽāĻžāύ āĻšāĻŦā§
$(-1)^{4} \cdot{ }^{10} \mathrm{C}_{4} \cdot 2^{2}=840$
Answer: (A) āĻĒāĻā§āĻāĻŽ āĻāĻŦāĻ 840
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4. āĻāĻāĻāĻŋ āĻŦāϏā§āϤā§āϰ āĻāĻžā§āĻž āĻāĻĒāϰā§āϰ āĻĻāĻŋāĻā§ āĻĒā§āϰāĻā§āώā§āĻĒāύ āĻāϰāϞ⧠āύāĻŋāϰā§āĻĻāĻŋāώā§āĻ āĻŦāĻŋāύā§āĻĻā§ P-āϤ⧠āĻĒā§āĻāĻžāϤ⧠t1 āϏāĻŽā§ āϞāĻžāĻā§āĨ¤ āϝāĻĻāĻŋ āĻāϰāĻ t2 āϏāĻŽā§ āĻĒāϰ āĻŦāϏā§āϤā§āĻāĻŋ āĻā§āĻŽāĻŋāϤ⧠āĻĒāϤāĻŋāϤ āĻšā§ āϤāĻŦā§ āĻāĻŖāĻžāĻāĻŋāϰ āϏāϰā§āĻŦā§āĻā§āĻ āĻāĻā§āĻāϤāĻž āĻšāĻŦā§
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(A) $\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
(B) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
(C) $\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}{ }^{2}\right)$
(D) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}{ }^{2}\right)$
Answer: (B) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
āĻŽā§āĻ āĻŦāĻŋāĻāϰāĻŖāĻāĻžāϞ = T = t1 + t2
āĻāĻŦāĻžāϰ,
$\mathrm{T}=\frac{2 \mathrm{u}}{\mathrm{g}}$
$\Rightarrow \mathrm{u}=\frac{1}{2} \mathrm{~g} \mathrm{~T}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\Rightarrow \mathrm{u}^{2}=\frac{1}{4} \mathrm{~g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
â´ āϏāϰā§āĻŦā§āĻā§āĻ āĻāĻā§āĻāϤāĻž āĻšāĻŦā§
$\frac{u^{2}}{2 g}=\frac{1}{8} g\left(t_{1}+t_{2}\right)^{2}$
Answer: $(\mathrm{B}) \frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
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5. $\tan \left(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right)$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦā§
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(A) $\frac{5}{6}$
(B) $1$
(C) $\frac{\pi}{4}$
(D) $-\frac{5}{6}$
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Answer: (B) $1$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
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āϏāĻžāϰāĻžāϏāϰāĻŋ Calculator āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻāϰā§:
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6. $\left|\begin{array}{lll}\mathrm{a} & 1 & \mathrm{~b}+\mathrm{c} \\ \mathrm{b} & 1 & \mathrm{c}+\mathrm{a} \\ \mathrm{c} & 1 & \mathrm{a}+\mathrm{b}\end{array}\right|$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦā§
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(A) 0
(B) $a b c(a+b)(b+c)(c+a)$
(C) abc
(D) $(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$
Answer: (A) o
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\left|\begin{array}{lll}\mathrm{a} & 1 & \mathrm{~b}+\mathrm{c} \\ \mathrm{b} & 1 & \mathrm{c}+\mathrm{a} \\ \mathrm{c} & 1 & \mathrm{a}+\mathrm{b}\end{array}\right|$
$=\left|\begin{array}{lll}a+b+c & 1 & b+c \\ a+b+c & 1 & c+a \\ a+b+c & 1 & a+b\end{array}\right|$
$=a+b+c\left|\begin{array}{lll}1 & 1 & b+c \\ 1 & 1 & c+a \\ 1 & 1 & a+b\end{array}\right| \quad\left[c_{1}^{\prime}=c_{1}+c_{3}\right]$
$=0$
Answer: (A) $0$
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7. 3P āĻāĻŦāĻ 2P āĻŦāϞāĻĻā§āĻŦā§ā§āϰ āϞāĻŦā§āϧāĻŋ RāĨ¤ āĻĒā§āϰāĻĨāĻŽ āĻŦāϞ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻāϰāϞ⧠āϞāĻŦā§āϧāĻŋāϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖāĻ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻšā§āĨ¤ āĻŦāϞāĻĻā§āĻŦā§ā§āϰ āĻ āύā§āϤāϰā§āĻāϤ āĻā§āĻŖ āĻšāĻŦā§
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(A) 110°
(B) 120°
(C) 150°
(D) 135°
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Answer: (B) 120°
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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āĻŽāύ⧠āĻāϰāĻŋ, 3P āĻāĻŦāĻ 2P āĻŽāĻžāύā§āϰ āĻŦāϞāĻĻā§āĻŦā§ā§āϰ āĻ āύā§āϤāϰā§āĻāϤ āĻā§āĻŖ ÎąāĨ¤
$\therefore \mathrm{R}^{2}=9 \mathrm{P}^{2}+4 \mathrm{P}^{2}+12 \mathrm{P}^{2} \cos \alpha=13 \mathrm{P}^{2}+12 \mathrm{P}^{2} \cos \alpha$
āĻĒā§āϰāĻĨāĻŽ āĻŦāϞā§āϰ āĻŽāĻžāύ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻāϰāϞā§
$4 \mathrm{R}^{2}=(6 \mathrm{P})^{2}+4 \mathrm{P}^{2}+2.2 \mathrm{P} .6 \mathrm{P} \cos \alpha=36 \mathrm{P}^{2}+4 \mathrm{P}^{2}+24 \mathrm{P}^{2} \cos \alpha=40 \mathrm{P}^{2}+24 \mathrm{P}^{2} \cos \alpha$
$\therefore \frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{P}^{2}(40+24 \cos \alpha)}{\mathrm{P}^{2}(13+12 \cos \alpha)}$
$\Rightarrow 52+48 \cos \alpha=40+24 \cos \alpha$
$\Rightarrow 24 \cos \alpha=-12$
$\Rightarrow \cos \alpha=-1 / 2$
$\therefore \alpha=120^{\circ}$
Answer: (B) $120^{\circ}$
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8. āĻāĻāĻāĻŋ āĻāϞā§āĻāĻā§āϰāĻŋāĻ āĻĢāĻŋāϞā§āĻĄā§ āĻāϞā§āĻāĻā§āϰāύā§āϰ āϤā§āĻŦāϰāĻŖ āĻāĻŦāĻ āĻļāĻā§āϤāĻŋ āϏāĻŽāĻžāύā§āĻĒāĻžāϤāĻŋāĻāĨ¤ $10^{-20} \mathrm{~N}$ āĻļāĻā§āϤāĻŋāϰ āĻāύā§āϝ āϤā§āĻŦāϰāĻŖ $10^{10} \frac{\mathrm{m}}{\mathrm{s}^{2}}$ āĻšāϞā§, $10^{-25} \mathrm{~N}$ āĻļāĻā§āϤāĻŋāϰ āĻāύā§āϝ āϤā§āĻŦāϰāĻŖ āĻšāĻŦā§
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(A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$
(B) $10^{15} \mathrm{~m} / \mathrm{s}^{2}$
(C) $10^{-5-} \mathrm{m} / \mathrm{s}^{2}$
(D) $10^{-15} \mathrm{~m} / \mathrm{s}^{2}$
Answer: (A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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āĻĻā§āĻā§āĻž āĻāĻā§, āϤā§āĻŦāϰāĻŖ â āĻļāĻā§āϤāĻŋ
$10^{10} \mathrm{~m} / \mathrm{s}^{2} \propto 10^{-20} \mathrm{~N}$
āĻāĻŦāĻžāϰ,
$x \propto 10^{-25}$
$\therefore \frac{x}{10^{10}}=\frac{10^{-25}}{10^{-20}}$
$\Rightarrow \mathrm{x}=10^{-25} \cdot 10^{10} \cdot 10^{20}=10^{5}$
Answer: (A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$
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9. āĻĻāĻļāĻŽāĻŋāĻ āϏāĻāĻā§āϝāĻž $2013$ āĻāϰ āĻĻā§āĻŦāĻŋāĻŽāĻŋāĻā§ āĻĒā§āϰāĻāĻžāĻļ āĻšāĻŦā§
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(A) 11111011101
(B) 10111011111
(C) 10101110111
(D) 10101110101
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Answer: (A) 11111011101
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$2 \mid 2012$
$2 \mid 1006-1$
$2 \mid 503-0$
$2 \mid 251-1$
$2 \mid 125-1$
$2 \mid 62-1$
$2 \mid 31-0$
$2 \mid 15-1$
$2 \mid 7-1$
$2 \mid 3-1$
$2 \mid 1-1$
 $0-1$
$\therefore(2013)_{10}=(11111011101)_{2}$
Answer: (A) 11111011101
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10. x = y2 āĻāĻŦāĻ y = x â 2 āĻĻā§āĻŦāĻžāϰāĻž āĻāĻŦāĻĻā§āϧ āĻā§āώā§āϤā§āϰā§āϰ āĻā§āώā§āϤā§āϰāĻĢāϞ āĻšāĻŦā§
(A) $1 \frac{1}{3}$
(B) $1 \frac{1}{3}$
(C) $4 \frac{1}{2}$
(D) $4 \frac{3}{4}$
Answer: (C) $4 \frac{1}{2}$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{x}=\mathrm{y}^{2}$
$\mathrm{y}=\mathrm{x}-2 \Rightarrow \mathrm{x}=\mathrm{y}+2$
$\therefore \mathrm{y}^{2}=\mathrm{y}+2$
$\Rightarrow \mathrm{y}^{2}-\mathrm{y}-2=0$
$\Rightarrow \mathrm{y}^{2}-2 \mathrm{y}+\mathrm{y}-2=0$
$\Rightarrow \mathrm{y}(\mathrm{y}-2)+1(\mathrm{y}-2)=0$
$\Rightarrow(\mathrm{y}-2)(\mathrm{y}+1)=0$
$\Rightarrow \mathrm{y}=-1,2$
$\int_{-1}^{2}\left(\mathrm{y}+2-\mathrm{y}^{2}\right) \mathrm{dy}$
$=\int_{-1}^{2} \mathrm{y} \mathrm{dy}+2 \int_{-1}^{2} \mathrm{dy}-\int_{-1}^{2} \mathrm{y}^{2} \mathrm{dy}$
$=\left[\frac{1}{2} y^{2}\right]_{-1}^{2}+[2 y]_{-1}^{2}-\left[\frac{1}{3} y^{3}\right]_{-1}^{2}$
$=\frac{1}{2}(4-1)+2(2+1)-\frac{1}{3}(8+1)$
$=4 \frac{1}{2}$
Answer: (C) $4 \frac{1}{2}$
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11. āϝāĻĻāĻŋ $y=\sqrt{\cos 2 x}$ āĻšā§, āϤāĻŦā§Â $\frac{d y}{d x}=$
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(A) $-\frac{\sin 2 x}{\sqrt{\cos 2 x}}$
(B) $-\frac{\cos 2 x}{\sqrt{\sin 2 x}}$
(C) $-\frac{2 \sin 2 x}{\sqrt{\tan x}}$
(D) $\frac{\tan 2 \mathrm{x}}{\sqrt{\sin 2 \mathrm{x}}}$
Answer: (A) $-\frac{\sin 2 x}{\sqrt{\cos 2 x}}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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$\mathrm{y}=\sqrt{\cos 2 \mathrm{x}}$ $\frac{d y}{d x}=\frac{1}{2 \sqrt{\cos 2 \mathrm{x}}} \frac{d}{d x} \cos 2 \mathrm{x}=\frac{1}{2 \sqrt{\cos 2 \mathrm{x}}}(-\sin 2 \mathrm{x}) \frac{d}{d x} 2 \mathrm{x}=-\frac{\sin 2 \mathrm{x}}{\sqrt{\cos 2 \mathrm{x}}}$ Answer: (A) $-\frac{\sin 2 \mathrm{x}}{\sqrt{\cos 2 \mathrm{x}}}$
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12. āĻāĻāĻāύ āĻā§āώāĻ āĻāĻāĻāĻŋ āĻā§āϤāĻžāĻāĻžāϰ āĻŦāĻžāĻāĻžāύā§āϰ āϤāĻŋāύ āĻĻāĻŋāĻ āĻŦā§ā§āĻž āĻĻāĻŋā§ā§ āĻāĻŦāĻ āĻāϤā§āϰā§āĻĻāĻŋāĻā§ āĻāĻāĻāĻŋ āĻĻā§āĻā§āĻžāϞ āĻĻāĻŋā§ā§ āĻā§āϰāĻžāĻ āĻĻāĻŋāϞāĨ¤ āϝāĻĻāĻŋ āϤāĻžāĻāϰ āĻāĻžāĻā§ $100 m$ āĻŦā§ā§āĻž āĻĨāĻžāĻā§ āϤāĻŦā§ āĻā§āϰāĻžāĻ āĻĻā§āĻā§āĻž āϏā§āĻĨāĻžāύā§āϰ āϏāϰā§āĻŦā§āĻā§āĻ āĻā§āϤāύ āĻšāĻŦā§
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(A)Â 2500 m2
(B)Â 1250 m2
(C)Â 750 m2
(D)Â 2000 m2
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Answer: (B) 1250 m2
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13. sin (ax + b) āĻāϰ n-āϤāĻŽ āĻ āύā§āϤāϰāĻ āĻšāĻŦā§
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(A) $a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$
(B) $a^{n} \cos \left(\frac{\pi}{2} n+a x+b\right)$
(C) $(-1)^{\mathrm{n}} a^{\mathrm{n}} \sin (a x+b)$
(D) $(-1)^{\mathrm{n}} \mathrm{a}^{\mathrm{n}} \cos (\mathrm{ax}+\mathrm{b})$
Answer: (A) $a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$y=\sin (a x+b)$
$y_{1}=a \cos (a x+b)=a \sin \left(\frac{\pi}{2}+a x+b\right)$
$y_{2}=a^{2} \cos \left(\frac{\pi}{2}+a x+b\right)=a^{2} \sin \left(2 \cdot \frac{\pi}{2}+a x+b\right)$
$y_{3}=a^{3} \cos \left(2 \cdot \frac{\pi}{2}+a x+b\right)=a^{3} \sin \left(3 \cdot \frac{\pi}{2}+a x+b\right)$
-------------------------------------------------
$\therefore \mathrm{y}_{\mathrm{n}}=\mathrm{a}^{\mathrm{n}} \sin \left(\frac{\pi}{2} \mathrm{n}+\mathrm{ax}+\mathrm{b}\right)$
Answer: $(A) a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$
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14. âaâ āĻāϰ āĻā§āύ āĻŽāĻžāύā§āϰ āĻāύā§āĻ¯Â $2 \hat{\imath}+\hat{\jmath}-\hat{k}, 3 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}$ āĻāĻŦāĻ $\hat{\imath}-3 \hat{\jmath}+a \hat{k}$ āĻā§āĻā§āĻāϰāϤā§āϰ⧠āϏāĻŽāϤāϞā§ā§?
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(A)Â 5
(B)Â 4
(C)Â 3
(D)Â 2
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Answer: (A)Â 5
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
  $\left|\begin{array}{ccc}2 & 1 & -1 \\ 3 & -2 & 4 \\ 1 & -3 & a\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ 11 & 2 & 4 \\ 2 a+1 & a-3 & a\end{array}\right|=0$
$\Rightarrow-(11 a-33-4 a-2)=0$
$\Rightarrow 7 a-35=0$
$\therefore a=5$
Answer: (A) 5
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15. $8+4 \sqrt{5} i$ āĻāϰ āĻŦāϰā§āĻāĻŽā§āϞ āĻšāĻŦā§
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$(\mathrm{A}) \pm(3-2 \mathrm{i})$
(B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$
(C) $\pm \sqrt{10}-\sqrt{2} \mathrm{i}$
(D) $\pm(3+2 i)$
Answer: (B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$a=8$
$b=4 \sqrt{5}$
$(8+4 \sqrt{5} i)^{\frac{1}{2}}$
$=\pm\left[\left\{\frac{1}{2}\left(\sqrt{a^{2}+b^{2}}+a\right)\right\}^{\frac{1}{2}}+i\left\{\frac{1}{2}\left(\sqrt{a^{2}+b^{2}}-a\right)\right\}^{\frac{1}{2}}\right]$
$=\pm\left\{\frac{1}{2}(8+\sqrt{64+80})\right\}^{\frac{1}{2}}+i\left\{\frac{1}{2}(-8+\sqrt{64+80})\right\}^{\frac{1}{2}}$
$=\pm\left\{\frac{1}{2}(8+12)\right\}^{\frac{1}{2}}+\mathrm{i}\left\{\frac{1}{2}(-8+12)\right\}^{\frac{1}{2}}$
$=\pm \sqrt{10}+\sqrt{2} \mathrm{i}$
Answer: (B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$
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16. āϝāĻĻāĻŋ $x^{2}+3 x y+5 y^{2}=1$ āĻšā§, āϤāĻžāĻšāϞ⧠$\frac{d y}{d x}$ āϏāĻŽāĻžāύ āĻšāĻŦā§
(A) $-\frac{2 x+3 y}{3 x+10 y}$
(B) $\frac{2 x+3 y}{3 x+10 y}$
(C) $\frac{2 x-3 y}{3 x+10 y}$
(D) $\frac{2 x+3 y}{3 x-10 y}$
Answer: (A) $-\frac{2 x+3 y}{3 x+10 y}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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$\mathrm{x}^{2}+3 \mathrm{xy}+5 \mathrm{y}^{2}=1$
$\Rightarrow 2 \mathrm{x}+3 \mathrm{y}+3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+10 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\Rightarrow \frac{d y}{d x}(3 x+10 y)=-(2 x+3 y)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2 \mathrm{x}+3 \mathrm{y}}{3 \mathrm{x}+10 \mathrm{y}}$
Answer: $(\mathrm{A})-\frac{2 \mathrm{x}+3 \mathrm{y}}{3 \mathrm{x}+10 \mathrm{y}}$
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17. $\frac{1}{2}+\frac{1}{3^{2}}+\frac{1}{2^{8}}+\frac{1}{3^{4}}+\frac{1}{2^{5}}+\frac{1}{3^{6}}+\cdots$ āϧāĻžāϰāĻžāϰ āϏāĻŽāώā§āĻāĻŋ āĻšāĻŦā§
(A) $\frac{24}{19}$
(B) $\frac{19}{24}$
(C) $\frac{5}{24}$
(D) $\frac{5}{19}$
Answer: (B) $\frac{19}{24}$
18. $x^{2}-5 x-3=0$ āϏāĻŽā§āĻāϰāĻŖā§āϰ āĻŽā§āϞāĻĻā§āĻŦā§ $Îą$, $β$ āĻšāϞ⧠$\frac{1}{\alpha}, \frac{1}{\beta}$ āĻŽā§āϞāĻŦāĻŋāĻļāĻŋāώā§āĻ āϏāĻŽā§āĻāϰāĻŖāĻāĻŋ āĻšāĻŦā§?
(A) $3 x^{2}-5 x+1=0$
(B) $x^{2}+5 x+3=0$
(C) $5 x^{2}-3 x-1=0$
(D) $3 \mathrm{x}^{2}+5 \mathrm{x}-1=0$
Answer: (D) $3 x^{2}+5 x-1=0$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
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$a x^{2}+b x+c=0$ āϏāĻŽā§āĻāϰāĻŖā§āϰ āĻŽā§āϞāĻĻā§āĻŦā§ $alpha$ āĻ $beta$ āĻšāϞā§Â $\frac{1}{\alpha}$ āĻ $\frac{1}{\beta}$ āĻŽā§āϞāĻŦāĻŋāĻļāĻŋāώā§āĻ āϏāĻŽā§āĻāϰāĻŖ āĻšāĻŦā§, $c x^{2}+b x+a=0$
Answer: (D) 3x2 + 5x â 1 = 0
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19. āĻŦāĻžāϏā§āϤāĻŦ āϏāĻāĻā§āϝāĻžā§Â $\frac{1}{|2 x-3|}>5$ āĻ āϏāĻŽāϤāĻžāĻāĻŋāϰ āϏāĻŽāĻžāϧāĻžāύ āĻšāϞā§
(A) $\left(\frac{7}{5}, \frac{8}{5}\right)$
(B) $\left[\frac{7}{5}, \frac{8}{5}\right]$
(C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$
(D) $\left[\frac{7}{5}, \frac{3}{2}\right] \cup\left[\frac{3}{2}, \frac{8}{5}\right]$
Answer: (C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$|2 x-3|<\frac{1}{5}$
$\Rightarrow-\frac{1}{5}<2 x-3<\frac{1}{5}$
$\Rightarrow \frac{7}{5}<x<\frac{8}{5}$
āĻāĻŦāĻžāϰ, $x=\frac{3}{2}$ āĻšāϞ⧠$\frac{1}{|2 x-3|}=\frac{1}{0}$ āĻ āϏāĻāĻā§āĻāĻžā§āĻŋāϤāĨ¤
Answer:Â (C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$
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20. āĻŽāĻžāύ āύāĻŋāϰā§āĻŖā§ āĻāϰ: $\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}$
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(A) 1
(B) $-1$
(C) 2
(D) 3
Answer: (A) 1
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
lâHôpitalâs rule āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻāϰ⧠āĻĒāĻžāĻ,
$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=\lim _{x \rightarrow 0} e^{x}=1$
Answer: (A) 1
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21. $\int \sqrt{\frac{1+x}{1-x}} d x=f(x)+c$ āĻšāϞā§, $\mathrm{f}(\mathrm{x})$ āĻāϰ āĻŽāĻžāύ
(A) $\sin ^{-1} x+\sqrt{1-x^{2}}$
(B) $\sin ^{-1} x-\sqrt{1-x^{2}}$
(C) $\cos ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$
(D) $\sin ^{-1} x-\sqrt{1+x^{2}}$
Answer: (B) $\sin ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\int \sqrt{\frac{1+x}{1-x}} d x$
$=\int \frac{1+x}{\sqrt{1-x^{2}}} d x$
$=\int \frac{d x}{\sqrt{1-x^{2}}}+=\int \frac{x d x}{\sqrt{1-x^{2}}}$
$=\sin ^{-1} x-\sqrt{1-x^{2}}$
Answer:Â (B) $\sin ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$
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22. $x^{2}+4 x+2 y=0$ āĻĒāϰāĻžāĻŦā§āϤā§āϤā§āϰ āĻļā§āϰā§āώāĻŦāĻŋāύā§āĻĻā§ āĻšāĻŦā§
(A) $(2,-2)$
(B) $(-2,-2)$
(C) $(-2,2)$
(D) $(2,2)$
Answer: (C) $(-2,2)$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{x}^{2}+4 \mathrm{x}+2 \mathrm{y}=0$
$\Rightarrow(\mathrm{x}+2)^{2}=-2(\mathrm{y}-2) \ldots \ldots \ldots(\mathrm{i})$
āϧāϰāĻŋ,
$\mathrm{x}+2=\mathrm{X}, \mathrm{y}-2=\mathrm{Y}$
$(\mathrm{i}) \Rightarrow \mathrm{X}^{2}=-2 \mathrm{Y}$
āĻļā§āϰā§āώāĻŦāĻŋāύā§āĻĻā§āϰ āϏā§āĻĨāĻžāύāĻžāĻā§āĻ: $\mathrm{X}=0, \mathrm{Y}=\mathrm{O}$
$\therefore \mathrm{x}=-2, \mathrm{y}=2$
Answer: (C) $(-2,2)$
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23. $f(x)=4-(x-3)^{2}$ āĻĢāĻžāĻāĻļāύā§āϰ āĻĄā§āĻŽā§āĻāύ āĻāĻŦāĻ āϰā§āĻāĻā§āĻ āϝāĻĨāĻžāĻā§āϰāĻŽā§
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(A) $\mathbf{R}, \mathbf{R}$
(B) $\mathbf{R}, \mathrm{x} \leq 4$
(C) $x \geq 4, \mathbf{R}$
(D) $\mathbf{R}, x \geq 3$
Answer: (B) $\mathbf{R}, x \leq 4$
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
$\mathrm{x}$ āĻāϰ āϏāĻŦ āĻŦāĻžāϏā§āϤāĻŦ āĻŽāĻžāύā§āϰ āĻāύā§āϝ $f(x)$ āĻāϰ āĻŦāĻžāϏā§āϤāĻŦ āĻŽāĻžāύ āĻĒāĻžāĻā§āĻž āϝāĻžā§āĨ¤
â´ āĻĢāĻžāĻāĻļāύā§āϰ āĻĄā§āĻŽā§āĻāύ = R
āϧāϰāĻŋ,
$y=f(x)=4-(x-3)^{2}$
$\Rightarrow(x-3)^{2}=4-y$
$\Rightarrow(x-3)=\pm \sqrt{4-y}$
$\Rightarrow x=3 \pm \sqrt{4-y}$
$\mathrm{y} \geq 4$ āĻšāϞā§Â $\mathrm{x}$ āĻāϰ āĻāĻāĻŋāϞ āĻŽāĻžāύ āĻĒāĻžāĻā§āĻž āϝāĻžā§ āϝāĻž āĻĄā§āĻŽā§āĻāύ āĻāϰ āϏāĻĻāϏā§āϝ āύā§āĨ¤
â´Â $\mathrm{x}$ āĻāϰ āϏāĻŦ āĻŦāĻžāϏā§āϤāĻŦ āĻŽāĻžāύā§āϰ āĻāύā§āĻ¯Â $y \leq 4$ āĻšāĻŦā§āĨ¤
â´ āĻĢāĻžāĻāĻļāύā§āϰ āϰā§āĻāĻā§āĻ = $\{x: x \in \mathbf{R}$ āĻāĻŦāĻ $x \leq 4\}$
Answer:Â (B) $\mathbf{R}, x \leq 4$
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24. $\mathrm{x}$-āĻ āĻā§āώāĻā§ $(4, 0)$ āĻŦāĻŋāύā§āĻĻā§āϤ⧠āϏā§āĻĒāϰā§āĻļ āĻāϰ⧠āĻāĻŦāĻ āĻā§āύā§āĻĻā§āĻ°Â $5 x-7 y+1=0$ āϏāϰāϞāϰā§āĻāĻžāϰ āĻāĻĒāϰ āĻ āĻŦāϏā§āĻĨāĻŋāϤ āĻāĻŽāύ āĻŦā§āϤā§āϤā§āϰ āϏāĻŽā§āĻāϰāĻŖ āĻšāĻŦā§
(A) $x^{2}+y^{2}-8 x-6 y+9=0$
(B) $x^{2}+y^{2}-8 x+6 y+16=0$
(C) $x^{2}+y^{2}-8 x+6 y+9=0$
(D) $x^{2}+y^{2}-8 x-6 y+16=0$
Answer: (B) $x^{2}+y^{2}-8 x+6 y+16=0$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
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āϧāϰāĻŋ, āĻŦā§āϤā§āϤā§āϰ āϏāĻŽā§āĻāϰāĻŖ, $x^{2}+y^{2}+2 g x+2 f y+c=0$
âĩ āĻŦā§āϤā§āϤāĻāĻŋ $\mathrm{x}$ āĻ āĻā§āώāĻā§ $(4,0)$ āĻŦāĻŋāύā§āĻĻā§āϤ⧠āϏā§āĻĒāϰā§āĻļ āĻāĻ°ā§ â´ āĻā§āύā§āĻĻā§āϰā§āϰ āĻā§āĻ $=4=-g$ āĻāĻŦāĻ $c=g^{2}=16$
âĩ āĻā§āύā§āĻĻā§āĻ°Â $5 x-7 y+1=0$ āϰā§āĻāĻžāϰ āĻāĻĒāϰ āĻ āĻŦāϏā§āĻĨāĻŋāĻ¤Â $\therefore 5(4)-7(-f)+1=0 \Rightarrow f=3$
â´ āĻŦā§āϤā§āϤā§āϰ āϏāĻŽā§āĻāϰāĻŖ, $x^{2}+y^{2}+2 g x+2 f y+c=0=x^{2}+y^{2}-8 x+6 y+16=0$
Answer:Â (B) $x^{2}+y^{2}-8 x+6 y+16=0$
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25. $A=\left[\begin{array}{cc}1 & \mathrm{i} \\ -\mathrm{i} & 1\end{array}\right], B=\left[\begin{array}{cc}\mathrm{i} & -1 \\ -1 & -\mathrm{i}\end{array}\right]$ āĻāĻŦāĻ $\mathrm{i}=\sqrt{-1}$ āĻšāϞ⧠$\mathrm{AB}$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦā§
(A) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
(C) $\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right]$
(D) $\left[\begin{array}{ll}\mathrm{i} & 1 \\ 1 & \mathrm{i}\end{array}\right]$
Answer: (B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\mathrm{AB}$
$=\left[\begin{array}{cc}1 & \mathrm{i} \\ -\mathrm{i} & 1\end{array}\right] \times\left[\begin{array}{cc}\mathrm{i} & -1 \\ -1 & -\mathrm{i}\end{array}\right]$
$=\left[\begin{array}{cc}\mathrm{i}-\mathrm{i} & -1-\mathrm{i}^{2} \\ -\mathrm{i}^{2}-1 & \mathrm{i}-\mathrm{i}\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Answer:Â (B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
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26. āϏā§āĻŦāϰāĻŦāϰā§āĻŖāĻā§āϞā§āĻā§ āϏāĻŦ āϏāĻŽā§ āĻāĻāϤā§āϰ⧠āϰā§āĻā§ KACHUA āĻļāĻŦā§āĻĻāĻāĻŋāϰ āĻŦāϰā§āĻŖāĻā§āϞā§āĻā§ āϏāĻžāĻāĻžāύā§āϰ āϏāĻāĻā§āϝāĻž āĻšāĻŦā§
Â
(A)Â 24
(B)Â 72
(C)Â 144
(D)Â 8
Â
Answer:Â (B)Â 72
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āϏāĻžāĻŽāĻžāϧāĻžāύ :
Â
KACHUA āĻļāĻŦā§āĻĻāĻāĻŋāϤ⧠āĻŽā§āĻ āĻŦāϰā§āĻŖ $6$āĻāĻŋ āϝāĻžāϰ $3$āĻāĻŋ āϏā§āĻŦāϰāĻŦāϰā§āĻŖāĨ¤ āϏā§āĻŦāϰāĻŦāϰā§āĻŖ 3āĻāĻŋ āĻā§ āĻāĻāĻāĻŋ āĻŦāϰā§āĻŖ āϧāϰāϞ⧠āĻŽā§āĻ āĻŦāϰā§āĻŖ āĻšā§ $(6 â 3 + 1) = 4$ āĻāĻŋ āϝāĻžāĻĻā§āϰāĻā§Â $4 !$ āĻāĻĒāĻžā§ā§ āϏāĻžāĻāĻžāύ⧠āϝāĻžā§āĨ¤ āĻāϰāĻžāϰ āϏā§āĻŦāϰāĻŦāϰā§āĻŖ 3āĻāĻŋ āĻā§ āύāĻŋāĻā§āĻĻā§āϰ āĻŽāϧā§āϝ⧠$\frac{3 !}{2 !}$ āĻāĻĒāĻžā§ā§ āϏāĻžāĻāĻžāύ⧠āϝāĻžā§āĨ¤
â´ āĻŽā§āĻ āĻŦāĻŋāύā§āϝāĻžāϏ āϏāĻāĻā§āϝāĻžÂ $=4 ! \times \frac{3 !}{2 !}=72$
Answer: (B) 72
Â
27. āĻāĻāĻāύ āϞā§āĻā§āϰ $3$ āĻā§ā§āĻž āĻāĻžāϞ⧠āĻŽā§āĻāĻž āĻāĻŦāĻ $2$ āĻā§ā§āĻž āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻž āĻāĻā§āĨ¤ āĻāĻāĻĻāĻŋāύ āĻ āύā§āϧāĻāĻžāϰ⧠āϤāĻžā§āĻžāĻšā§ā§āĻž āĻāϰ⧠āϞā§āĻāĻāĻŋ āĻāĻžāĻĒā§ āĻĒāϰāϞāĨ¤ āϏ⧠āĻĒā§āϰāĻĨāĻŽā§ āĻāĻāĻāĻŋ āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻž āĻĒāϰāĻžāϰ āĻĒāϰ āĻĒāϰāĻŦāϰā§āϤ⧠āĻŽā§āĻāĻžāĻ āĻŦāĻžāĻĻāĻžāĻŽā§ āĻšāĻā§āĻžāϰ āϏāĻŽā§āĻāĻžāĻŦāύāĻž
(A) $\frac{1}{3}$
(B) $\frac{2}{15}$
(C) $\frac{1}{10}$
(D) $\frac{3}{10}$
Answer: (C) $\frac{2}{15}$
Â
āϏāĻžāĻŽāĻžāϧāĻžāύ :
                   āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻžāϰ āϏāĻāĻā§āϝāĻž = 2Ã2 = 4
                   āĻāĻžāϞā§āĻž āĻŽā§āĻāĻžāϰ āϏāĻāĻā§āϝāĻž = 2Ã3 = 6
āĻĒā§āϰāĻĨāĻŽā§ āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻž āĻĒāϰāĻžāϰ āϏāĻŽā§āĻāĻŦā§āϝāϤāĻžÂ $=\frac{4}{10}=\frac{2}{5}$
āĻĒāϰ⧠āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻž āĻĒāϰāĻžāϰ āϏāĻŽā§āĻāĻŦā§āϝāϤāĻžÂ $=\frac{3}{9}=\frac{1}{3}$
āĻĒā§āϰāĻĨāĻŽā§ āĻāĻāĻāĻŋ āĻŦāĻžāĻĻāĻžāĻŽā§ āĻŽā§āĻāĻž āĻĒāϰāĻžāϰ āĻĒāϰ āĻĒāϰāĻŦāϰā§āϤ⧠āĻŽā§āĻāĻžāĻ āĻŦāĻžāĻĻāĻžāĻŽā§ āĻšāĻā§āĻžāϰ āϏāĻŽā§āĻāĻžāĻŦāύāĻžÂ $\frac{2}{5} \times \frac{1}{3}=\frac{2}{15}$
Answer:Â (C) $\frac{2}{15}$
Â
28. $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-3}{2 \mathrm{x}+1}$ āĻāĻŦāĻ $\mathrm{x} \neq \frac{1}{2}$ āĻšāϞ⧠$\mathrm{f}^{-1}(-2)$ āĻāϰ āĻŽāĻžāύ āĻāϤ āĻšāĻŦā§
(A) $\frac{1}{2}$
(B) $\frac{1}{5}$
(C) 2
(D) 5
Answer: (B) $\frac{1}{5}$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
āϧāϰāĻŋ,
$\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-3}{2 \mathrm{x}+1}$
$\Rightarrow 2 \mathrm{xy}+\mathrm{y}=\mathrm{x}-3$
$\Rightarrow 2 \mathrm{xy}-\mathrm{x}=-\mathrm{y}-3$
$\Rightarrow \mathrm{x}=\frac{\mathrm{y}+3}{1-2 \mathrm{y}}$
$\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}+3}{1-2 \mathrm{x}}$
$f^{-1}(-2)=\frac{-2+3}{1-2(-2)}=\frac{1}{5}$
Answer: (B) $\frac{1}{5}$
29. $\mathrm{u}$ āĻŦā§āĻā§ āĻ āύā§āĻā§āĻŽāĻŋāĻā§āϰ āϏāĻžāĻĨā§ $alpha$ āĻā§āĻŖā§ āĻĒā§āϰāĻā§āώāĻŋāĻĒā§āϤ āĻŦāϏā§āϤā§āϰ āϏāϰā§āĻŦā§āĻā§āĻ āĻāĻā§āĻāϤāĻž āĻšāĻŦā§
(A) $\frac{\mathrm{u}^{2} \sin 2 \alpha}{2 \mathrm{~g}}$
(B) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}$
(C) $\frac{\mathrm{u}^{2} \sin 2 \alpha}{\mathrm{g}}$
(D) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{\mathrm{g}}$
Answer: (B) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}$
Â
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30. $\frac{(i+1)^{2}}{(i-1)^{4}}$ āĻāĻāĻŋāϞ āϏāĻāĻā§āϝāĻžāĻāĻŋāϰ āĻāϰā§āĻā§āĻŽā§āύā§āĻ āĻšāĻŦā§
(A)Â Ī
(B)Â â Ī
(C)Â Â $\frac{\pi}{2}$
(D) â $\frac{\pi}{2}$
Â
Answer: (D) â $\frac{\pi}{2}$
āϏāĻžāĻŽāĻžāϧāĻžāύ :
$\frac{(i+1)^{2}}{(i-1)^{4}}$
$=\frac{(i+1)^{2}}{(i-1)^{2}(i-1)^{2}}$
$=\frac{\mathrm{i}^{2}+2 \mathrm{i}+1}{\left(\mathrm{i}^{2}-2 \mathrm{i}+1\right)\left(\mathrm{i}^{2}-2 \mathrm{i}+1\right)}$
$=\frac{2 \mathrm{i}}{(-2 \mathrm{i})(-2 \mathrm{i})}$
$=\frac{2 \mathrm{i}}{4 \mathrm{i}^{2}}$
$=-\frac{1}{2} \mathrm{i}$
â´ āĻāĻāĻŋāϞ āϏāĻāĻā§āϝāĻžāĻāĻŋāϰ āĻāϰā§āĻā§āĻŽā§āύā§āĻ $-\frac{\pi}{2}$
Answer: (D) â $\frac{\pi}{2}$
Â
āϰāϏāĻžā§āύ
Â
Â
1. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āĻā§ā§āĻžāύā§āĻāĻžāĻŽ āϏā§āĻāĻāĻŋ āĻĒāϰāĻŽāĻžāύā§āϰ āĻāĻāĻāĻŋ āĻāϞā§āĻāĻā§āϰāύā§āϰ āĻāύā§āϝ āϏāĻŽā§āĻāĻŦ āύ⧠?
(A) $\mathrm{n}=2, \mathrm{I}=1, \mathrm{~m}=\mathrm{o}, \mathrm{s}=+1 / 2$
(B) $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=2, \mathrm{~s}=-1 / 2$
(C) $\mathrm{n}=1, \mathrm{I}=0, \mathrm{~m}=0, \mathrm{~s}=-1 / 2$
(D) $n=2, \mathrm{I}=0, \mathrm{~m}=0, \mathrm{~s}=+1 / 2$
Â
āĻāϤā§āϤāϰ : (B) $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=2, \mathrm{~s}=-1 / 2$
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2. āĻšāĻžāĻāĻĄā§āϰā§āĻā§āύā§āϰ āĻĒāĻžāϰāĻŽāĻžāĻŖāĻŦāĻŋāĻ āĻŦāϰā§āĻŖāĻžāϞā§āϰ āĻā§āύ āϏāĻŋāϰāĻŋāĻāĻāĻŋāϤ⧠āĻĻā§āĻļā§āϝāĻŽāĻžāύ āĻ āĻā§āĻāϞā§āϰ āϰāĻļā§āĻŽāĻŋ āĻĻā§āĻāĻž āϝāĻžā§ ?
Â
(A) Paschen
(B)Â Lyman
(C)Â Balmer
(D)Â Brackett
Â
āĻāϤā§āϤāϰ : (C) Balmer
Â
3. $10.0g$ āĻ āĻā§āϏāĻŋāĻā§āύ⧠āĻ āĻŖā§āϰ āϏāĻāĻā§āϝāĻž āĻāϤ ?
Â
(A) $3.76 \times 10^{23}$
(B) $6.02 \times 10^{22}$
(C) $9.63 \times 10^{23}$
(D) $1.88 \times 10^{23}$
Â
āĻāϤā§āϤāϰ : (D) $1.88 \times 10^{23}$
Â
4. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύāĻāĻŋāϰ āĻŦā§āϝāĻžāϏāĻžāϰā§āϧ āϏāĻŦāĻā§ā§ā§ āĻāĻŽ ?
Â
(A) $_{9} \mathrm{~F}^{-}$
(B) ${ }_{10} \mathrm{Ne}$
(C) ${ }_{11} \mathrm{Na}^{+}$
(D) $12 \mathrm{Mg}^{2+}$
Â
āĻāϤā§āϤāϰ : (A) $_{9} \mathrm{~F}^{-}$
Â
5. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āϝā§āĻāĻāĻŋ āϏāĻŦāĻā§ā§ā§ āĻāĻŽ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžā§ āĻŦāĻŋā§ā§āĻāĻŋāϤ āĻšāĻŦā§ ?
Â
(A) $\mathrm{Na}_{2} \mathrm{CO}_{3}$
(B) $\mathrm{K}_{2} \mathrm{CO}_{3}$
(C) $\mathrm{MgCO}_{3}$
(D) $\mathrm{BaCO}_{3}$
Â
āĻāϤā§āϤāϰ : (C) $\mathrm{MgCO}_{3}$
Â
6.āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āϝā§āĻāĻāĻŋāϰ āĻāĻā§āϤāĻŋ āϤā§āϰāĻŋāĻā§āĻŖāĻžāĻāĻžāϰ āϏāĻŽāϤāϞā§ā§ ?
Â
(A) $\mathrm{BCl}_{3}$
(B) $\mathrm{H}_{3} \mathrm{O}^{+}$
(C) $\mathrm{BrF}_{5}$
(D) $\mathrm{PH}_{3}$
Â
āĻāϤā§āϤāϰ : (A) $\mathrm{BCl}_{3}$
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7. $300 k$ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžā§ āĻŦāĻžāϤāĻžāϏā§āϰ $\mathrm{N}_{2}$ āĻ āĻŖā§āϰ āĻāϤāĻŋ āĻāϤ ?
(A) $450 \mathrm{~m} / \mathrm{s}$
(B) $516 \mathrm{~m} / \mathrm{s}$
(C) $400 \mathrm{~m} / \mathrm{s}$
(D) $600 \mathrm{~m} / \mathrm{s}$
Â
āĻāϤā§āϤāϰ : (B) $516 \mathrm{~m} / \mathrm{s}$
Â
8. āĻā§āϰāĻŋāĻāύāĻžāϰāĻĄ āĻŦāĻŋāĻāĻžāϰāĻ āĻšāϞ⧠â
(A) CH3ONa
(B) RBaCl
(C) RMgX
(D) RCaX
Â
Â
āĻāϤā§āϤāϰ : (C) RMgX
Â
9. āĻāĻĨāĻžāύāϞā§āĻ°Â $170^{\circ} \mathrm{C}$ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžā§ āĻ āϤāĻŋāϰāĻŋāĻā§āϤ āĻāĻžā§ āϏāĻžāϞāĻĢāĻŋāĻāϰāĻŋāĻ āĻāϏāĻŋāĻĄ āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻā§āϰāĻŋā§āĻž āĻāϰāĻžāϞ⧠āĻā§ āĻā§āĻĒāύā§āύ āĻšā§ ?
(A) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SO}_{4}$
(B) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{SO}_{4}$
(C) $\mathrm{CH}_{2}=\mathrm{CH}_{2}$
(D) $\mathrm{CH}_{3} \mathrm{CHO}$
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āĻāϤā§āϤāϰ : (C) $\mathrm{CH}_{2}=\mathrm{CH}_{2}$
10. $10.0$ āĻŽāĻŋ.āϞāĻŋ. $\mathrm{NaOH}$ āĻĻā§āϰāĻŦāĻŖāĻā§ $0.12 \mathrm{M}$ āĻāύāĻŽāĻžāϤā§āϰāĻžā§ $15.0$ āĻŽāĻŋ.āϞāĻŋ āĻ āĻā§āϏāĻžāϞāĻŋāĻ āĻāϏāĻŋāĻĄ āĻĻā§āĻŦāĻžāϰāĻž āĻāĻžāĻāĻā§āϰā§āĻļāύ āĻāϰāϞ⧠āĻĒā§āϰāĻļāĻŽāύ āĻŦāĻŋāύā§āĻĻā§ āĻĒāĻžāĻā§āĻž āϝāĻžā§ āĨ¤Â $\mathrm{NaOH}$ āĻāϰ āĻāύāĻŽāĻžāϤā§āϰāĻž āĻāϤ ?
(A) $0.25 \mathrm{M}$
(B) $0.36 \mathrm{M}$
(C) $0.32 \mathrm{M}$
(D) $0.40 \mathrm{M}$
Â
āĻāϤā§āϤāϰ : (B) $0.36 \mathrm{M}$
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11.āĻŽā§āϝāĻžāύā§āϏāĻŋā§āĻžāĻŽ āĻĢāϏāĻĢā§āĻā§āϰ āϏāĻāĻā§āϤ āĻšāϞ⧠-
Â
(A) $\mathrm{Mg}_{2}\left(\mathrm{PO}_{4}\right)_{2}$
(B) $\mathrm{MgPO}_{4}$
(C) $\mathrm{Mg}_{2}\left(\mathrm{PO}_{4}\right)_{2}$
(D) $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
Â
āĻāϤā§āϤāϰ : (D) $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
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12. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āĻŦāĻŋāĻā§āϰāĻŋā§āĻžāĻāĻŋ āĻĄāĻŋāϏāĻĒā§āϰā§āĻĒāϰāύā§āĻļāύ āĻŦāĻŋāĻā§āϰāĻŋā§āĻž ?
Â
(A) $\mathrm{H}_{2} \mathrm{~S}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}+\mathrm{S}$
(B) $\mathrm{CuSO}_{4}+\mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2}+\left(\mathrm{NH}_{3}\right)_{2} \mathrm{SO}_{4}$
(C) $\mathrm{Fe}+$ dil. $\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{FeSO}_{4}+\mathrm{H}_{2}$
(D) $\mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
Â
āĻāϤā§āϤāϰ : (D) $\mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
Â
13.$18.5 \% \mathrm{~N}_{2} \mathrm{O}_{4}$ $25^{\circ} \mathrm{C}$ āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžā§ āĻāĻŦāĻ $1 \mathrm{~atm}$ āĻāĻžāĻĒā§ āĻŦāĻŋā§ā§āĻāĻŋāϤ āĻšāϞ⧠$K_{p}$ āĻāϰ āĻŽāĻžāύ āĻāϤ ?
Â
(A) $0.142 \mathrm{~atm}$
(B) $0.185 \mathrm{~atm}$
(C) $0.220 \mathrm{~atm}$
(D) $0.125 \mathrm{~atm}$
Â
āĻāϤā§āϤāϰ : (A) $0.142 \mathrm{~atm}$
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Â
14. āύāĻŋāĻŽā§āύā§āϰ āĻŽāĻŋāĻļā§āϰāĻŖāϏāĻŽā§āĻš āĻĨā§āĻā§ āĻŦāĻžāĻĢāĻžāϰ āĻĻā§āϰāĻŦāĻŖāĻāĻŋ āĻļāύāĻžāĻā§āϤ āĻāϰ -
Â
(A) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$
(B) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$
(C) $0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$
(D) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{HCl}+0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$
Â
āĻāϤā§āϤāϰ : (A) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$
Â
Â
15. āύāĻŋāĻā§āϰ āĻā§āύ āϝā§āĻāĻāĻŋ Fehling āĻĻā§āϰāĻŦāĻŖā§āϰ āϏāĻžāĻĨā§ āĻŦāĻŋāĻā§āϰāĻŋā§āĻž āĻāϰ⧠āϞāĻžāϞ āĻ āϧāĻāĻā§āώā§āĻĒ āĻĻā§ā§ ?
Â
(A) $\mathrm{RCH}_{2} \mathrm{X}$
(B) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}$
(C) $\mathrm{RCH}_{2} \mathrm{OH}$
(D) $\mathrm{RCH}_{2} \mathrm{CHO}$
Â
āĻāϤā§āϤāϰ : (D) $\mathrm{RCH}_{2} \mathrm{CHO}$
Â
16. āĻāĻāĻāĻŋ $\mathrm{CH}_{3} \mathrm{CN}$ āĻ āĻŖā§āϤā§Â $\pi$ āĻāĻŦāĻ $\sigma$ āĻŦāύā§āϧāύā§āϰ āϏāĻāĻā§āϝāĻž āϝāĻĨāĻžāĻā§āϰāĻŽā§ -
(A) 5 and 2
(B) 4 and 3
(C) 5 and 3
(D) 4 and 2
Â
āĻāϤā§āϤāϰ : (A) 5 and 2
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17. $\mathrm{IUPAC}$ āύāĻžāĻŽāĻāϰāĻŖ āĻ āύā§āϏāĻžāϰ⧠$\mathrm{CH}_{3}-\mathrm{CH}\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)-\mathrm{CH}_{2}-\mathrm{CHBr}-\mathrm{CHCl}-\mathrm{CH}_{3}$ āĻāϰ āύāĻžāĻŽ āĻšāϞ⧠-
Â
(A)Â 2-chloro-3-bromo-5-ethylhexane
(B)Â 2-chloro-3-bromo-5-methyl heptane
(C)Â 3-bromo-2-chloro-5-ethyl-hexane
(D)Â 3-bromo-2-chloro-5-methyl heptane
Â
āĻāϤā§āϤāϰ : (B) 2-chloro-3-bromo-5-methyl heptane
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18. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύāĻāĻŋ āϏāĻŦāĻā§ā§ā§ āϏā§āĻĨāĻŋāϤāĻŋāĻļā§āϞ āĻāĻžāϰā§āĻŦā§-āĻā§āϝāĻžāĻāĻžā§āύ ?
Â
(A) $\mathrm{CH}_{3}{ }^{+}$
(B) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}^{+} \mathrm{H}$
(C) $\mathrm{H}_{2} \mathrm{C}^{+}-\mathrm{CH}_{3}$
(D) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$
Â
āĻāϤā§āϤāϰ : (D) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$
Â
19. āĻĒāĻāĻžāϏāĻŋā§āĻžāĻŽ āϏāĻžā§āĻžāύāĻžāĻāĻĄā§āϰ āĻāĻĒāϏā§āĻĨāĻŋāϤāĻŋāϤ⧠āĻšāĻžāĻāĻĄā§āϰā§āĻā§āύ āϏāĻžā§āĻžāύāĻžāĻāĻĄ āĻĒā§āϰāĻžāĻĒāĻžāύāϞā§āϰ āϏāĻžāĻĨā§ āĻŦāĻŋāĻā§āϰāĻŋā§āĻž āĻāϰ⧠āĨ¤āĻŦāĻŋāĻā§āϰāĻŋā§āĻžāϰ āĻā§āĻļāϞāĻāĻŋ āĻšāϞ⧠-
Â
(A)Â nuclophilic addition
(B)Â electrophilic substitution
(C)Â nucleophilic substituton
(D)Â electrophilic addition
Â
āĻāϤā§āϤāϰ : (C) nucleophilic substituton
Â
Â
20. āĻ ā§āϝāĻžāϰā§āĻŽā§āĻāĻŋāĻ āĻĒā§āϰāϤāĻŋāϏā§āĻĨāĻžāĻĒāύ āĻŦāĻŋāĻā§āϰāĻŋā§āĻžā§ āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āĻā§āϰā§āĻĒāĻāĻŋ āĻŽā§āĻāĻž āύāĻŋāϰā§āĻĻā§āĻļāĻ ?
(A) $-\mathrm{NO}_{3}$
(B) $-\mathrm{OH}$
(C) $-\mathrm{Cl}$
(D) $-\mathrm{CH}_{3}$
Â
āĻāϤā§āϤāϰ : (A) $-\mathrm{NO}_{3}$
Â
21. āϏā§āĻĄāĻž āĻ ā§āϝāĻžāĻļ āĻļāĻŋāϞā§āĻĒāĻā§āώā§āϤā§āϰ⧠āĻā§āύ āĻĒāĻĻā§āϧāϤāĻŋāϤ⧠āϤā§āϰāĻŋ āĻāϰāĻž āĻšā§ ?
Â
(A)Â Bessemer process
(B)Â Chamber process
(C)Â Solvary process
(D)Â Haber process
Â
āĻāϤā§āϤāϰ : (C) Solvary process
Â
22. āĻāĻŦāĻ āĻ āϧāĻāĻā§āώā§āĻĒ āϏāĻŽā§āĻšā§āϰ āϰāĻ-āĻāϰ āĻā§āϰāĻŽ āĻšāϞ⧠â
Â
(A)brown, pink, white and blue
(B)brown, blue white and pink
(C)pink, white , brown and blue
(D) brown, white, blue and pink
Â
āĻāϤā§āϤāϰ : (B)brown, blue white and pink
Â
23. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋ āĻ āϏāϤā§āϝ ?
(A) $\mathrm{NH}_{4}^{+}$is the conjugate acid of base $\mathrm{NH}_{3}$
(B) $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair
(C) $\mathrm{OH}$ is the conjugate base of acid $\mathrm{H}_{2} \mathrm{O}$
(D) $\mathrm{OH}^{-}$and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair
Â
āĻāϤā§āϤāϰ : (B) $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair
Â
24. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āϝā§āĻāĻāĻŋ āĻāϞā§ā§ āĻĻā§āϰāĻŦāĻŖā§ āϏāĻŦāĻā§ā§ā§ āϏāĻšāĻā§ āĻšāĻžāĻāĻĄā§āϰā§-āĻŦāĻŋāĻļā§āϞā§āώāĻŋāϤ āĻšā§ ?
(A) $\mathrm{CCl}_{4}$
(B) $\mathrm{SnCl}_{2}$
(C) $\mathrm{SiCl}_{4}$
(D) $\mathrm{PbCl}_{4}$
Â
āĻāϤā§āϤāϰ : (C) $\mathrm{SiCl}_{4}$
Â
25. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āϝā§āĻāĻāĻŋāϤ⧠āϏāĻā§āĻāĻžāϞāύ āĻ āĻā§āώāĻŽ $Ī$âāĻāϞā§āĻā§āĻā§āϰāύ āĻāĻ ?
Â
(A) $\mathrm{C}_{2} \mathrm{H}_{6}$
(B) $\mathrm{C}_{6} \mathrm{H}_{6}$
(C) $\mathrm{C}_{3} \mathrm{H}_{8}$
(D) $\mathrm{C}_{2} \mathrm{H}_{4}$
Â
āĻāϤā§āϤāϰ : (D) $\mathrm{C}_{2} \mathrm{H}_{4}$
Â
26. $\mathrm{Sr}, \mathrm{Tc}, \mathrm{Zr}$ āĻāĻŦāĻ $\mathrm{Rb}$ āĻĒāϰāĻŽāĻžāύā§āϰ āĻŦā§āϝāĻžāϏāĻžāϰā§āϧā§āϰ āĻā§āϰāĻŽ āĻšāϞ⧠â
Â
(A) $\mathrm{Rb}>\mathrm{Sr}>\mathrm{Zr}>\mathrm{Tc}$
(B) $\mathrm{Tc}>\mathrm{Sr}>\mathrm{Rb}>\mathrm{Zr}$
(C) $\mathrm{Sr}>\mathrm{Tc}>\mathrm{Zr}>\mathrm{Rb}$
(D) $\mathrm{Zr}>\mathrm{Tc}>\mathrm{Rb}>\mathrm{Sr}$
Â
āĻāϤā§āϤāϰ : (A) $\mathrm{Rb}>\mathrm{Sr}>\mathrm{Zr}>\mathrm{Tc}$
Â
Â
27. āύāĻŋāĻŽā§āύā§āϰ āĻā§āύ āĻŦāĻŋāĻā§āϰāĻŋā§āĻžāĻāĻŋ āĻĒā§āϰāĻļāĻŽāύ āĻāύāĻĨāĻžāϞāĻĒāĻŋ, $\Delta H_{\text {neutr }}$ āĻĒā§āϰāĻāĻžāĻļ āĻāϰ⧠?
Â
(A) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (aq) $+\mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow \mathrm{CaSO}_{4}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (aq) $+2 \mathrm{NH}_{3}$ (aq) $\rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ (aq)
(C) $\mathrm{HCl}$ (aq) $+1^{1 / 2} \mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow 1 / 2 \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)
(D) $2 \mathrm{HCl}$ (aq) $+\mathrm{Ca}$ (OH) $_{2}$ (aq) $\rightarrow \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)
Â
āĻāϤā§āϤāϰ : (C) $\mathrm{HCl}$ (aq) $+1^{1 / 2} \mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow 1 / 2 \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)
Â
28. $\mathrm{Zn}^{+} \mid \mathrm{Zn}$ āĻāĻŦāĻ $\mathrm{Ag}^{+} \mid \mathrm{Ag}$ āϤā§āĻŋā§āĻĻā§āĻŦāĻžāϰ āĻĻā§āĻāĻŋāϰ āĻŦāĻŋāĻāĻžāϰāĻŖ āĻŦāĻŋāĻāĻŦ āϝāĻĨāĻžāĻā§āϰāĻŽā§Â $-0.76 \mathrm{~V}$ āĻāĻŦāĻ $+0.80 \mathrm{~V}$ āĻāĻ āϤā§āĻŋā§āĻĻā§āĻŦāĻžāϰ āĻĻā§āĻāĻŋ āĻĻā§āĻŦāĻžāϰāĻž āϤā§āϰāĻŋ āĻā§āώā§āϰ āĻŽā§āĻ āĻŦāĻŋāĻāĻŦ āĻāϤ āĻšāĻŦā§ ?
Â
(A) $-0.04 \mathrm{~V}$
(B) $+1.56 \mathrm{~V}$
(C) +0.14 V
(D) $-1.56 \mathrm{~V}$
Â
āĻāϤā§āϤāϰ : (B) $+1.56 \mathrm{~V}$
Â
29.āĻŦā§āϞā§āĻāĻāĻŽā§āϝāĻžāύ āϧā§āϰā§āĻŦāĻā§āϰ āĻāĻāĻ āĻšāϞ⧠â
(A) J /molecule
(B) J.s
(C) $\mathrm{J} / \mathrm{K}$
(D) g/cc
Â
āĻāϤā§āϤāϰ : (A) J /molecule
Â
30. $\operatorname{Sn}(s)+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \operatorname{Sn}^{+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ - āĻŦāĻŋāĻā§āϰāĻŋā§āĻžāĻāĻŋāϰ āĻā§āώā§āϤā§āϰ⧠āύāĻŋāĻŽā§āύā§āϰ āĻā§āύāĻāĻŋ āĻā§āώā§āϰ āĻŦā§āĻĻā§āϧāĻŋ āĻāϰāĻŦā§ ?
Â
(A) icrease in the size of silver rod
(B) increase in the concentration of Sn+ ions
(C) increase in the concentration of Ag+ ions
(D) increase in the size of tin rod
Â
āĻāϤā§āϤāϰ : (C) increase in the concentration of Ag+ ions
Â
āĻā§āĻŦāĻŦāĻŋāĻā§āĻāĻžāύ
Â
ā§§. āĻŦāĻžāϏā§āĻ āϤāύā§āϤā§āϰ āĻā§āύāĻāĻŋ ?
      āĻ. āĻĒāĻžāĻ āϤāύā§āϤā§āϰ
      āĻ. āĻāĻžāϰā§āĻĒāĻžāϏ āϤāύā§āϤā§āϰ
      āĻ. āĻļāĻŋāĻŽā§āϞ āϤā§āϞāĻž
      āĻ. āĻā§ā§āϰ
Â
āĻāϤā§āϤāϰ : (āĻ) āĻĒāĻžāĻ āϤāύā§āϤā§āϰ
Â
⧍. āĻŽāĻžāύā§āώā§āϰ āĻŦāĻā§āώāĻĻā§āĻļā§ā§ āĻāĻļā§āϰā§āĻāĻž āĻā§āĻāĻŋ ?
      āĻ. 7
      āĻ. 12
      āĻ. 10
      āĻ. 15
Â
āĻāϤā§āϤāϰ : āĻ. 12
Â
ā§Š. āĻā§āύāĻāĻŋ ommatidium āĻāϰ āĻ āĻāĻļ āύ⧠?
Â
      āĻ. rhabdosome
      āĻ. rentinal sheath
      āĻ. retinal cell
      āĻ. ocellus
Â
āĻāϤā§āϤāϰ : āĻ. ocellus
Â
ā§Ē. āĻā§āĻĻāϰā§āĻ āϏā§āώā§āĻāĻŋāĻāĻžāϰ⧠āĻĒāϰāĻā§āĻŦāĻŋāϰ āύāĻžāĻŽ-
      āĻ. entamoeba histolytica
      āĻ. Wuchereria bancrofti
      āĻ. Ades fatigans
      āĻ. Culex quinquefasciatus
Â
āĻāϤā§āϤāϰ : āĻ. Wuchereria bancrofti
Â
ā§Ģ. Liliopsida āĻŦāϞāϤ⧠āĻāĻŋ āĻŦā§āĻāĻžā§?
      āĻ. āĻāĻāĻŦā§āĻāĻĒāϤā§āϰ⧠āĻāĻĻā§āĻāĻŋāĻĻ
      āĻ. āĻĻā§āĻŦāĻŋāĻŦā§āĻāĻĒāϤā§āϰ⧠āĻāĻĻā§āĻāĻŋāĻĻ
      āĻ. āϞāĻŋāϞāĻŋā§ā§āϏ⧠āĻā§āϤā§āϰ
      āĻ. āϞāĻŋāϞāĻŋā§ā§āϞāĻŋāϏ āĻŦāϰā§āĻ
Â
āĻāϤā§āϤāϰ : āĻ. āĻāĻāĻŦā§āĻāĻĒāϤā§āϰ⧠āĻāĻĻā§āĻāĻŋāĻĻ
Â
ā§Ŧ. āĻāϰāĻļā§āϞāĻžāϰ āĻĒā§āϰāĻŋāĻā§āϰāĻĢāĻŋāĻ āĻĒāϰā§āĻĻāĻž āĻĨāĻžāĻā§ āĻā§āύ āϏā§āĻĨāĻžāύā§?
      āĻ. āĻā§āϰāĻĒ
      āĻ. āĻāĻŋāĻāĻžāϰā§āĻĄ
      āĻ. āĻšā§āĻĒāĻžāĻāĻŋāĻ āϏāĻŋāĻāĻžāĻŽ
      āĻ. āĻŽā§āϏā§āύāĻāĻŋāϰāύ
Â
āĻāϤā§āϤāϰ :  āĻ. āĻŽā§āϏā§āύāĻāĻŋāϰāύ
Â
ā§. F1-āĻāύā§āϰ āĻāĻĻā§āĻāĻŋāĻĻāĻā§ āĻĒā§āϰāĻā§āĻāύā§āύ āĻĒā§āϝāĻžāϰā§āύā§āĻā§āϰ āϏāĻžāĻĨā§ āĻā§āϰāϏ āĻāϰāĻžāĻā§ āĻŦāϞāĻž āĻšā§
      āĻ. āĻŦā§āϝāĻžāĻ āĻā§āϰāϏ
      āĻ. āĻā§āϏā§āĻ āĻā§āϰāϏ
      āĻ. āĻŽāύā§āĻšāĻžāĻāĻŦā§āϰāĻŋāĻĄ āĻā§āϰāϏ
      āĻ. āĻĄāĻžāĻāĻšāĻžāĻāĻŦā§āϰāĻŋāĻĄ āĻā§āϰāϏ
Â
āĻāϤā§āϤāϰ : āĻ. āĻā§āϏā§āĻ āĻā§āϰāϏ
Â
ā§Ž. āĻŽāĻžāύā§āώā§āϰ āĻĻā§āĻšā§ āĻā§āύāĻāĻŋ āĻā§āϏā§āĻā§āϏā§āĻā§āϰāύ āϤā§āϰāĻŋ āĻāϰā§?
Â
      āĻ. āϏā§āĻā§āϰā§āĻāĻžāĻŽ
      āĻ. āĻāύā§āĻāĻžāϰāĻāĻŋāϏā§āĻāĻŋāĻļāĻŋā§āĻžāϞ āĻā§āώ
      āĻ. āĻāĻĒāĻŋāĻĄāĻŋāĻĄāĻžāĻāĻŽāĻŋāϏ
      āĻ. āϏā§āĻĒāĻžāϰāĻŽā§āĻā§āĻā§āύāĻŋā§āĻž
Â
āĻāϤā§āϤāϰ :  āĻ. āĻāύā§āĻāĻžāϰāĻāĻŋāϏā§āĻāĻŋāĻļāĻŋā§āĻžāϞ āĻā§āώ
Â
Â
⧝. āĻā§āϰā§āĻŦāϏ āĻāĻā§āϰ⧠āĻāϤāĻāĻŋ NADH2 āϤā§āϰāĻŋ āĻšā§?
      āĻ. 1
      āĻ. 2
      āĻ. 3
      āĻ. 4
Â
āĻāϤā§āϤāϰ : āĻ. 3
Â
ā§§ā§Ļ. āĻā§āϞāĻžāϰā§ā§āĻĄ āĻŽā§āϞ āĻā§āĻĨāĻžā§ āĻĒāĻžāĻā§āĻž āϝāĻžā§?
Â
      āĻ. Cycas
      āĻ. Ficus
      āĻ. Hibiscus
      āĻ. Daucus
Â
āĻāϤā§āϤāϰ : āĻ. Cycas
Â
ā§§ā§§. āĻāύāϏā§āϞāĻŋāύ āύāĻŋāĻāϏāϰāĻŖāĻāĻžāϰ⧠āĻā§āϰāύā§āĻĨāĻŋāϰ āύāĻžāĻŽ āĻšāϞ-
      āĻ. āϝāĻā§āϤ
      āĻ. āĻ āĻā§āύā§āϝāĻžāĻļā§
      āĻ. āĻĒā§āϞā§āĻšāĻž
      āĻ. āĻāĻāϞā§āĻāϏ āĻ āĻŦ āϞā§āϝāĻžāĻā§āĻāĻžāϰāĻšā§āϝāĻžāύā§āϏ
Â
āĻāϤā§āϤāϰ : āĻ. āĻāĻāϞā§āĻāϏ āĻ āĻŦ āϞā§āϝāĻžāĻā§āĻāĻžāϰāĻšā§āϝāĻžāύā§āϏ
Â
⧧⧍. āĻā§āύāĻāĻŋ āĻā§āϞ⧠āĻŽā§āϞā§āĻĄ?
      āĻ. Penicillium
      āĻ. Saprolegnia
      āĻ. Agaricus
      āĻ. Helminthosporium
Â
āĻāϤā§āϤāϰ : āĻ. Penicillium
Â
ā§§ā§Š. āĻ āϏā§āĻĨāĻŋ āϝ⧠āĻāĻŦāϰāĻŖ āĻĻā§āĻŦāĻžāϰāĻž āĻāĻŦā§āϤ āĻĨāĻžāĻā§ āϤāĻžāĻā§ āĻŦāϞā§-
      āĻ. āĻĒā§āϰāĻŋāĻā§āύāĻŋā§āĻžāĻŽ
      āĻ. āĻĒā§āϰāĻŋāĻ āϏā§āĻāĻŋā§āĻžāĻŽ
      āĻ. āĻĒā§āϰāĻŋāĻāĻžāϰā§āĻĄāĻŋā§āĻžāĻŽ
      āĻ. āĻĒā§āϰāĻŋāĻāύā§āĻĄā§āϰāĻŋā§āĻžāĻŽ
Â
āĻāϤā§āϤāϰ : āĻ. āĻĒā§āϰāĻŋāĻ āϏā§āĻāĻŋā§āĻžāĻŽ
Â
ā§§ā§Ē. āĻāĻŋāύā§āĻā§āϰ āĻā§āϞāĻā§āϰ āϏāĻŦāĻā§ā§ā§ āĻāĻŋāϤāϰā§āϰ āϏā§āϤāϰā§āϰ āύāĻžāĻŽ āĻšāϞ-
      āĻ. āύā§āϝāĻžāĻāĻžāϰ āϏā§āϤāϰ
      āĻ. āĻĒā§āϰāĻŋāĻāĻŽā§āϝāĻžāĻāĻŋāĻ āϏā§āϤāϰ
      āĻ. āĻĒā§āϰāĻŋāĻāϏā§āĻā§āϰāĻžāĻāĻžāĻŽ
      āĻ. āĻŽā§āϝāĻžāύā§āĻāϞ
Â
Â
āĻāϤā§āϤāϰ : āĻ. āύā§āϝāĻžāĻāĻžāϰ āϏā§āϤāϰ
Â
ā§§ā§Ģ. āύāĻŋāĻāĻā§āϞāĻŋā§āĻžāϏā§āϰ āĻĒā§āϰāĻĨāĻŽ āĻŦāϰā§āĻŖāύāĻž āĻāϰā§āύ-
     āĻ. āϰāĻŦāĻžāϰā§āĻ āĻŦā§āϰāĻžāĻāύ
      āĻ. āĻ ā§āϝāĻžāϰāĻŋāϏā§āĻāĻāϞ
      āĻ. āϞā§āĻ āĻĒāĻžāϏā§āϤā§āϰ
      āĻ. āĻāϞā§āĻāĻāĻžāύā§āĻĄāĻžāϰ āĻĢā§āϞā§āĻŽāĻŋāĻ
Â
āĻāϤā§āϤāϰ : āĻ. āϰāĻŦāĻžāϰā§āĻ āĻŦā§āϰāĻžāĻāύ
Â
ā§§ā§Ŧ. āĻĢā§āϏāĻĢā§āϏā§ā§ āϧāĻŽāύ⧠āĻŦāĻšāύ āĻāϰā§-
      āĻ. oxygenetaed blood
      āĻ. deoxygenated blood
      āĻ. pure blood
      āĻ. venous blood
Â
āĻāϤā§āϤāϰ : āĻ. deoxygenated blood
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ā§§ā§. āĻāĻāϞāĻŋā§āĻž āĻšāϞ-
      āĻ. āĻĻā§āώā§āĻāĻŋāϰ āϏāĻžāĻĨā§ āϏāĻāĻļā§āϞāĻŋāώā§āĻ āĻ āĻā§āĻ
      āĻ. āĻļā§āϰāĻŦāĻŖā§āϰ āϏāĻžāĻĨā§ āϏāĻāĻļā§āϞāĻŋāώā§āĻ āĻ āĻā§āĻ
      āĻ. āĻā§āϰāĻžāĻŖā§āϰ āϏāĻžāĻĨā§ āϏāĻāĻļā§āϞāĻŋāώā§āĻ āĻ āĻā§āĻ
      āĻ. āĻāĻžāĻĻā§āϝ āĻāϰā§āĻŦāĻŖā§āϰ āϏāĻžāĻĨā§ āϏāĻāĻļā§āϞāĻŋāώā§āĻ āĻ āĻā§āĻ
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āĻāϤā§āϤāϰ : āĻ. āĻļā§āϰāĻŦāĻŖā§āϰ āϏāĻžāĻĨā§ āϏāĻāĻļā§āϞāĻŋāώā§āĻ āĻ āĻā§āĻ
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ā§§ā§Ž. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋ āĻā§āĻŦāϞāĻŽāĻžāϤā§āϰ āĻŦāĻžāĻāϞāĻžāĻĻā§āĻļā§ āĻĒāĻžāĻā§āĻž āϝāĻžā§?
      āĻ. Knema bengalensis
      āĻ. Tactaria chattagramica
      āĻ. Artocarpus heterophyllus
      āĻ. Ficus benghalensis
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āĻāϤā§āϤāϰ : āĻ. Knema bengalensis
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⧧⧝. 'Diversity and classification of flowering plants' āĻŦāĻāĻāĻŋāϰ āϞā§āĻāĻ āĻā§?
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      āĻ. Carolus Linnaeus
      āĻ. George Bentham
      āĻ. Theophrastus
      āĻ. Armen Takhtajan
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āĻāϤā§āϤāϰ : āĻ. Armen Takhtajan
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⧍ā§Ļ. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋāϤ⧠āĻĒā§āϞāĻžāϏāĻŽāĻŋāĻĄ āύā§āĻ?
      āĻ. E. coli
      āĻ. A. Tumefaciens
      āĻ. Yeast
      āĻ. Chlorella
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āĻāϤā§āϤāϰ : āĻ. Yeast
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⧍⧧. āĻŽāĻžāύā§āώā§āϰ RBC āĻāϰ āĻā§ āĻā§ā§āώā§āĻāĻžāϞ āĻšāĻā§āĻā§-
      āĻ. 96 days
      āĻ. 120 days
      āĻ. 28 days
      āĻ. 62 days
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āĻāϤā§āϤāϰ : āĻ. 120 days
Â
⧍⧍. āĻā§āύāĻāĻŋ āϏāĻžā§āĻžāύā§āĻŦā§āϝāĻžāĻāĻā§āϰāĻŋā§āĻž āύā§?
      āĻ. Nostoc
      āĻ. Anabaena
      āĻ. Aulosira
      āĻ. Clostridium
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āĻāϤā§āϤāϰ : āĻ. Clostridium
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ā§¨ā§Š. āϏāĻĒā§āϤāĻŽ āĻāϰā§āĻāĻŋāĻāĻž āϏā§āύāĻžā§ā§āĻā§ āĻŦāϞāĻž āĻšā§-
      āĻ. āĻā§āĻāĻžāϏ
      āĻ. āĻā§āϰāĻā§āϞāĻŋā§āĻžāϰ
      āĻ. āĻĢā§āϝāĻžāϏāĻŋā§āĻžāϞ
      āĻ. āĻ āĻĒāĻāĻŋāĻ
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āĻāϤā§āϤāϰ : āĻ. āĻĢā§āϝāĻžāϏāĻŋā§āĻžāϞ
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⧍ā§Ē. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋāĻā§ āĻšāĻŋāϞ āĻŦāĻŋāĻā§āϰāĻŋā§āĻž āĻŦāϞāĻž āĻšā§?
      āĻ. CO2 āĻĨā§āĻā§ O2 āύāĻŋāϰā§āĻāĻŽāύ
      āĻ. H2O āĻĨā§āĻā§ O2 āύāĻŋāϰā§āĻāĻŽāύ
      āĻ. S2O āĻĨā§āĻā§ O2 āύāĻŋāϰā§āĻāĻŽāύ
      āĻ. NO2 āĻĨā§āĻā§ O2 āύāĻŋāϰā§āĻāĻŽāύ
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āĻāϤā§āϤāϰ : āĻ. H2O āĻĨā§āĻā§ O2 āύāĻŋāϰā§āĻāĻŽāύ
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⧍ā§Ģ. āĻāϞā§āĻāĻžāϰā§āϰ āĻāĻžāĻ āĻā§?
      āĻ. āĻāĻžāĻĻā§āϝ āϤā§āϰ⧠āĻāϰāĻž
      āĻ. āĻāĻžāĻĻā§āϝ āϏāĻā§āĻā§ āĻāϰāĻž
      āĻ. āϏā§āĻĒā§āϰ āύāĻŋāϰā§āĻāĻŽāύ āϏāĻžāĻšāĻžāϝā§āϝ āĻāϰāĻž
      āĻ. āĻŦāĻāĻļ āĻŦāĻŋāϏā§āϤāĻžāϰ⧠āĻ āĻāĻļāĻā§āϰāĻšāĻŖ āĻāϰāĻž
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āĻāϤā§āϤāϰ : āĻ. āϏā§āĻĒā§āϰ āύāĻŋāϰā§āĻāĻŽāύ āϏāĻžāĻšāĻžāϝā§āϝ āĻāϰāĻž
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⧍ā§Ŧ. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋ āĻĢāĻžāϰā§āĻŖ āĻĒāĻžāϤāĻžāϰ āύāĻžāĻŽ?
      āĻ. Stomium
      āĻ. Fronds
      āĻ. Prothallus
      āĻ. Scale leave
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āĻāϤā§āϤāϰ : āĻ. Fronds
Â
⧍ā§. āĻŽā§āϏā§āĻā§āϞāĻŋā§āĻž āĻĒāĻžāĻā§āĻž āϝāĻžā§ āĻā§āύ āĻĒā§āϰāĻžāĻŖā§āϤā§?
      āĻ. āĻŽāĻžāύā§āώ
      āĻ. āĻāϰāĻļā§āϞāĻž
      āĻ. āĻšāĻžāĻāĻĄā§āϰāĻž
      āĻ. āĻŽāĻžāĻ
Â
āĻāϤā§āϤāϰ : āĻ. āĻšāĻžāĻāĻĄā§āϰāĻž
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ā§¨ā§Ž. T2 āĻŦā§āϝāĻžāĻāĻā§āϰāĻŋāĻāĻĢāĻžā§ā§ āĻā§āĻāĻŋ āĻā§āύ āĻĨāĻžāĻā§?
      āĻ. 150
      āĻ. 145
      āĻ. 155
      āĻ. 160
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āĻāϤā§āϤāϰ :
Â
⧍⧝. āĻā§āĻžāϞā§āϏā§āϰ āϞāĻžāĻāύ āĻšāϞ⧠āĻāĻāĻāĻŋ āĻāĻžāϞā§āĻĒāύāĻŋāĻ āϏā§āĻŽāĻžāϰā§āĻāĻž ............ āĻāĻŦāĻ .......... āĻ āĻā§āĻāϞā§āϰ āĻŽāϧā§āϝ⧠āĻ āĻŦāϏā§āĻĨāĻŋāϤāĨ¤
     āĻ. āĻāϰāĻŋā§ā§āύā§āĻ āĻ āĻŦāĻžāĻāϞāĻžāĻĻā§āĻļ
      āĻ. āĻāĻāϰā§āĻĒ āĻ āĻāĻļāĻŋā§āĻž
      āĻ. āĻŦāĻžāĻāϞāĻžāĻĻā§āĻļ āĻ āĻāĻžāϰāϤ
      āĻ. āĻā§āϰāĻŋā§āĻž āĻ āĻāĻžāĻĒāĻžāύ
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āĻāϤā§āϤāϰ : āĻ. āĻāϰāĻŋā§ā§āύā§āĻ āĻ āĻŦāĻžāĻāϞāĻžāĻĻā§āĻļ
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ā§Šā§Ļ. āĻā§āύāĻāĻŋ āĻāĻžāĻāϞā§āĻŽ āĻāĻŋāϏā§āϝā§āϰ āĻ āĻāĻļ āύā§?
      āĻ. āϏāĻŋāĻ āύāϞ
      āĻ. āĻā§āϰā§āϝāĻžāĻāĻŋāĻĄ
      āĻ. āĻā§āϰā§āϝāĻžāĻāĻŋā§āĻž
      āĻ. āĻāĻĄ āĻĢāĻžāĻāĻŦāĻžāϰ
Â
āĻāϤā§āϤāϰ : āĻ. āϏāĻŋāĻ āύāϞ
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English
1. The synonym of 'embellish' is
(A) adorn
(B) shock
(C) impoverish
(D) destroy
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Answer: (A) adorn
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2. The verb of 'beauty' is
(A) beautician
(B) beautify
(C) beautiful
(D) beautification
Â
Answer: (B) beautify
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3. The idiom 'let things slide' means-
(A) ignore
(B) lose gradually
(C) reveal a secret
(D) set free
Â
Answer: (A) ignore
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4. Choose the correct spelling
(A) indegenus
(B) indiginous
(C) indigenous
(D) indigeneous
Â
Answer: (C) indigenous
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5. Why is the poet so sad to see the daffodils in 'The Daffodils'?
(A) The poet is sad because the flowers have not bloomed fully.
(B) The poet is sad because winter will come soon.
(C) The poet is sad because the flowers remind him of his own death.
(D) The poet is sad because flowers bring very hot weather.
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Answer: (C) The poet is sad because the flowers remind him of his own death.
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6. The correct translation of  âāϰāĻŦāĻŋāĻŦāĻžāϰ āĻšāĻāϤ⧠āĻŦā§āώā§āĻāĻŋ āĻšāĻāϤā§āĻāĻŋāϞâ-
(A) It was raining from Sunday.
(B) It has been raining from Sunday.
(C) It had been raining since Sunday.
(D) It rained since Sunday.
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Answer: (C) It had been raining since Sunday.
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7. The word 'constraint' means
(A) freedom
(B) limitaion
(C) plentiful
(D) endless
Â
Answer: (B) limitaion
Â
8. The expression 'to look after' means
(A) to take care
(B) to follow
(C) to imitate
(D) to gaze
Â
Answer: (A) to take care
Â
9. The antonym of âstubbornâ is
(A) agreeable                   (B) obstinate                    (C) difficult                       (D) irritable
Â
Answer: (A) agreeable
Â
10. What is the meaning of the expression âbottom of my heartâ?
(A) close to my heart
(B) core of my heart
(C) lower part of my heart
(D) close to my liver
Â
Answer: (B) core of my heart
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11. The word âdesperationâ is a/an
(A) adjective                    (B) verb                             (C) adverb                         (D) noun
Â
Answer: noun
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12. Why were the daffodils in Wordswothâs âI Wandered Lonely as a Cloudâ dancing?
(A) The poet was day dreaming.
(B) The flowers had cheerful company.
(C) The sea waves beside them had gone wild.
(D) There was a strong wind.
Â
Answer: (D) There was a strong wind.
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Questions 13-22: Fill in the blank/blanks:
13. Iâd like ------ information, please.
(A) an
(B) some
(C) few
(D) piece
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Answer: (B) some
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14. Yoga is ------ a good exercise for ------ breathing.
(A) taken, controlling
(B) treated, considerable
(C) not, slowly
(D) considered, controlling
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Answer: (D) considered, controlling
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15. A man ----- by a speeding bus while he was crossing the road.
(A) was run over
(B) was run down
(C) had been run
(D) has been run over
Â
Answer: (A) was run over
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16. He ----- prefers ----- speak very little.
(A) doesnât, to
(B) himself, to
(C) himself, for
(D) does, for
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Answer: (B) himself, to
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17. It ----- heavily when he ----- up.
(A) has snowed, woke
(B) snows, wake
(C) was snowing, woke
(D) is snowing, wakes
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Answer: (C) was snowing, woke
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18. Misuse of ----- energy has ----- destruction.
(A) solar, shown
(B) renewable, increase
(C) nuclear, cause
(D) atomic, wreaked
Â
Answer: (D) atomic, wreaked
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19. There is ----- milk in the bottle.
(A) very few
(B) any
(C) very little
(D) many
Â
Answer: (C) very little
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20. I have ------ him to give ----- smoking.
(A) said, up
(B) talked, for
(C) told, up
(D) told, in
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Answer: (C) told, up
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21. The groom arrived at the community centre exactly ----- time.
(A) in
(B) for
(C) by
(D) on
Â
Answer: (D) on
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22. He hates ----- kept ------.
(A) to be, waiting
(B) being kept, wait
(C) to be, to wait
(D) to, waiting
Â
Answer: (A) to be, waiting
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Questions 23-25: Choose the correct sentence:
23.
(A) It is you who is to pay.
(B) It is you who are to leave.
(C) It is you who is late.
(D) It is you who has won the prize.
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Answer: (B) It is you who are to leave.
Â
24.
(A) He suspicioned that something wrong.
(B) He suspicious that something is wrong.
(C) He suspected that something was wrong.
(D) He suspect that something is wrong.
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Answer: (C) He suspected that something was wrong.
25.
(A) The jury are arguing among themselves.
(B) The jury is arguing among themselves.
(C) The jury has argued among themselves.
(D) The jury has been arguing among themselves.
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Answer: (A) The jury are arguing among themselves.
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Read the passage and answer questions 26-30:
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Cats are carnivorous mammals of the family Felidae. They cannot chew their food, and their teeth are adapted to stab, anchor, and cut flesh. All cats except the cheetah have strong, sharp, retractile claws. They are not adapted for long chases but prowl their prey on padded feet and try to overwhelm it in short dash or pounce. Big cats roam over a large area, usually alone but sometimes in family groups, for example, a pride of lions can contain as many as 37 individuals. Cats generally are nocturnal animals, the retina of their eyes made extra sensitive to light by a layer of guanine, which causes the eyes to shine in the dark. Tigers are largest of the cats. They are identified by their characteristic striped coat. They inhabit forests and grasslands in Asia where populations have suffered from haunting, deforestation, and demand for tiger parts in traditional medicine.
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26. âCats are nocturnal animalsâ means
(A) cats have predatory eyesight
(B) cats are sensitive to smell
(C) cats haunt during daytime
(D) cats are active at night
Â
Answer: (D) cats are active at night
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27. A group of lions is called âpride of lionsâ, what is a group of dogs called?
(A) a pack of dogs
(B) a colony of dogs
(C) a flock of dogs
(D) a cluster of dogs
Â
Answer: (A) a pack of dogs
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28. âretractile clawsâ in the passage means
(A) claws the are extra sharp
(B) claws that can change shape
(C) claws that can move inwards
(D) claws that are sensitive to feelings
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Answer: (C) claws that can move inwards
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29. Cats are carnivorous, but human beings are
(A) herbivorous
(B) omnivorous
(C) farinaceous
(D) cadaverous
Â
Answer: (B) omnivorous
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30. The synonym of âprowlâ is
(A) dive
(B) chase
(C) ornace
(D) stalk
Â
Answer: (D) stalk
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āĻŦāĻžāĻāϞāĻž
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ā§§. āĻĒāĻžāĻā§āĻā§āϰā§- āĻāĻŦāĻŋāϤāĻžā§ āϝāĻžāϤā§āϰā§āϰāĻž āĻā§āĻĨāĻžā§ āĻŦāϏ⧠āĻ āĻĒā§āĻā§āώāĻž āĻāϰā§?
      āĻ. āϤā§āϰā§
      āĻ. āĻŽāĻžāϏā§āϤā§āϞā§
      āĻ. āĻŦāύā§āĻĻāϰā§
      āĻ. āĻĻāĻžāĻā§ā§
Â
āĻāϤā§āϤāϰ : āĻ. āĻŦāύā§āĻĻāϰā§
Â
āĻā§āĻāĻž: āϰāĻžāϤ āĻĒā§āĻšāĻžāĻŦāĻžāϰ āĻāϤ āĻĻā§āϰāĻŋ āĻĒāĻžāĻā§āĻā§āϰāĻŋ?/ āĻŦāύā§āĻĻāϰ⧠āĻŦāϏ⧠āϝāĻžāϤā§āϰā§āϰāĻž āĻĻāĻŋāύ āĻā§āύā§, (āϤā§āϤā§ā§ āϏā§āϤāĻŦāĻ)
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⧍. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋ āĻļā§āĻĻā§āϧ āĻŦāĻžāύāĻžāύ?
      āĻ. āϏā§āĻŦāĻžā§āϤā§āϤā§āĻŦāĻļāĻžāϏāύ
      āĻ. āϏāĻžā§āϤā§āϤā§āĻŦāĻļāĻžāϏāύ
      āĻ. āϏā§āĻŦāĻžā§āϤā§āϤāĻļāĻžāϏāύ
      āĻ. āϏā§āĻŦāĻžā§āϤā§āϤā§āĻŦāĻļāĻžāώāĻŖ
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āĻāϤā§āϤāϰ : āĻ. āϏā§āĻŦāĻžā§āϤā§āϤāĻļāĻžāϏāύ
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ā§Š. āĻŦāĻŋāϞāĻžāϏā§- āĻāϞā§āĻĒā§ āĻāύāĻŋāĻļ āĻļāϤāĻā§āϰ āϝ⧠āϏāĻŽāĻžāĻ-āϏāĻāϏā§āĻāĻžāϰā§āϰ āĻāĻĨāĻž āĻāĻā§ āϤāĻžāĻāϰ āύāĻžāĻŽ-
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      āĻ. āĻāĻļā§āĻŦāϰāĻāύā§āĻĻā§āϰ āĻŦāĻŋāĻĻā§āϝāĻžāϏāĻžāĻāϰ
      āĻ. āϰāĻžāĻŽāĻŽā§āĻšāύ āϰāĻžā§
      āĻ. āĻ āĻā§āώā§āĻā§āĻŽāĻžāϰ āĻĻāϤā§āϤ
      āĻ. āĻā§āĻĻā§āĻŦ āĻŽā§āĻā§āĻĒāĻžāϧā§āϝāĻžā§
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āĻāϤā§āϤāϰ : āĻ. āĻā§āĻĻā§āĻŦ āĻŽā§āĻā§āĻĒāĻžāϧā§āϝāĻžā§
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āĻā§āĻāĻž: āĻāĻŽāĻŋ āĻā§āĻĻā§āĻŦāĻŦāĻžāĻŦā§āϰ āĻĒāĻžāϰāĻŋāĻŦāĻžāϰāĻŋāĻ āĻĒā§āϰāĻŦāύā§āϧā§āϰāĻ āĻĻā§āώ āĻĻāĻŋāĻŦ āύāĻž āĻāĻŦāĻ āĻļāĻžāϏā§āϤā§āϰā§ā§ āĻŦāĻŋāϧāĻŋ-āĻŦā§āϝāĻŦāϏā§āĻĨāĻžāϰāĻ āύāĻŋāύā§āĻĻāĻž āĻāϰāĻŋāĻŦ āύāĻžāĨ¤ (āĻļā§āώ āĻ āύā§āĻā§āĻā§āĻĻ/āĻĒā§āϝāĻžāϰāĻžāĻā§āϰāĻžāĻĢ)
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ā§Ē. āϏāύā§āϧāĻŋāĻāĻāĻŋāϤ āĻā§āύ āĻļāĻŦā§āĻĻāĻāĻŋ āĻļā§āĻĻā§āϧ?
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  āĻ. āĻŦā§āĻšāĻĻāĻāĻļ
      āĻ. āĻāĻžāϤā§āϝāĻžāĻāĻŋāĻŽāĻžāύ
      āĻ. āĻāĻĻā§āϝāĻžāύā§āϤ
      āĻ. āĻļāĻŋāϰā§āĻā§āĻā§āĻĻ
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āĻāϤā§āϤāϰ : āĻ. āĻŦā§āĻšāĻĻāĻāĻļ
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ā§Ģ. āϤāĻžāĻŽā§āϰāĻļāĻžāϏāύ- āĻŽāĻžāύā§?
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      āĻ. āϏā§āĻŦā§āϰāĻļāĻžāϏāύ
      āĻ. āĻāĻžāϞ⧠āĻāĻāύ
      āĻ. āϤāĻžāĻŽāĻžāϰ āĻĒāĻžāϤ⧠āĻā§āĻĻāĻžāĻ āĻāϰāĻž āĻāĻĻā§āĻļ
      āĻ. āĻāĻāĻāύ āϰāĻžāĻāĻžāϰ āύāĻžāĻŽ
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āĻāϤā§āϤāϰ : āĻ. āϤāĻžāĻŽāĻžāϰ āĻĒāĻžāϤ⧠āĻā§āĻĻāĻžāĻ āĻāϰāĻž āĻāĻĻā§āĻļ
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ā§Ŧ. āĻĻā§āĻšā§āϰ āĻŦā§āϏ āĻ āύā§āϝāĻžā§ā§ āĻŽāύā§āϰ āĻŦā§āϏ āύāĻž āĻŦāĻžā§āϞ⧠āϤāĻžāĻā§ āĻŦāϞā§-
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      āĻ. āĻļāĻžāϰā§āϰāĻŋāĻ āĻĒā§āϰāϤāĻŋāĻŦāύā§āϧā§
      āĻ. āĻŽāύā§āĻŦāĻŋāĻāĻžāϰāĻā§āϰāϏā§āϤ
      āĻ. āĻŽāĻžāύāϏāĻŋāĻ āϰā§āĻā§
      āĻ. āĻŦā§āĻĻā§āϧāĻŋāĻĒā§āϰāϤāĻŋāĻŦāύā§āϧā§
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āĻāϤā§āϤāϰ : āĻ. āĻŦā§āĻĻā§āϧāĻŋāĻĒā§āϰāϤāĻŋāĻŦāύā§āϧā§
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ā§. āĻā§āύ āĻļāĻŦā§āĻĻāĻāĻŋ āĻļā§āĻĻā§āϧ?
      āĻ. āϏāĻŽā§āĻā§āώāύ
      āĻ. āϏāĻŽā§āĻā§āύ
      āĻ. āĻļāĻŋāϰāĻā§āĻā§āĻĻ
      āĻ. āĻāĻžāώā§āĻāϰ
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āĻāϤā§āϤāϰ : āĻ. āϏāĻŽā§āĻā§āύ
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ā§Ž. āĻŽāĻžāϤā§āĻšā§āĻĻā§ā§ āĻĒāĻā§āώāĻĒāĻžāϤāĻŋāϤāĻž āύāĻžāĻāĨ¤â āĻŦāĻžāĻā§āϝāĻāĻŋ āĻā§āύ āϰāĻāύāĻžāϰ āĻ āύā§āϤāϰā§āĻāϤ?
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      āĻ. āĻ āϰā§āϧāĻžāĻā§āĻā§
      āĻ. āϝā§āĻŦāύā§āϰ āĻāĻžāύ
      āĻ. āĻŦāĻŋāϞāĻžāϏā§
      āĻ. āĻāĻŽāϞāĻžāĻāĻžāύā§āϤā§āϰ āĻĻāĻĒā§āϤāϰ
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āĻāϤā§āϤāϰ : āĻ. āĻ āϰā§āϧāĻžāĻā§āĻā§
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⧝. āĻŦāĻžāĻāϞāĻž āĻŦā§āϝāĻā§āĻāύāĻŦāϰā§āĻŖā§āϰ āĻĒā§āϰāϤāĻŋāĻŦāϰā§āĻā§āϰ āĻĒāĻā§āĻāĻŽ āĻŦāϰā§āĻā§āϰ āϧā§āĻŦāύāĻŋāĻāĻŋ-
      āĻ. āĻā§āώāϧā§āĻŦāύāĻŋ
      āĻ. āĻ āĻā§āώāϧā§āĻŦāύāĻŋ
      āĻ. āĻŽāĻšāĻžāĻĒā§āϰāĻžāĻŖāϧā§āĻŦāύāĻŋ
      āĻ. āύāĻžāϏāĻŋāĻā§āϝāϧā§āĻŦāύāĻŋ
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āĻāϤā§āϤāϰ : āĻ. āύāĻžāϏāĻŋāĻā§āϝāϧā§āĻŦāύāĻŋ
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āϞāĻŋāĻā§āĻ: āϧā§āĻŦāύāĻŋ āĻ āĻŦāϰā§āĻŖ āĻĒā§āϰāĻāϰāĻŖ āĻ āĻāĻā§āĻāĻžāϰāĻŖāĻŦāĻŋāϧāĻŋ
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ā§§ā§Ļ. āϝā§āĻŽāύ āĻāϰā§āĻŽ āϤā§āĻŽāύ āĻĢāϞ- āĻ āĻŦāĻžāĻā§āϝ⧠āĻŦā§āϝāĻŦāĻšā§āϤ āĻšā§ā§āĻā§-
      āĻ. āύāĻŋāϰā§āϧāĻžāϰāĻ āĻŦāĻŋāĻļā§āώāĻŖ
      āĻ. āĻā§āϰāĻŋā§āĻž āĻŦāĻŋāĻļā§āώāĻŖ
      āĻ. āϏāĻžāĻĒā§āĻā§āώ āϏāϰā§āĻŦāύāĻžāĻŽ
      āĻ. āĻŦāĻŋāĻļā§āώāĻŖā§āϰ āĻŦāĻŋāĻļā§āώāĻŖ
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āĻāϤā§āϤāϰ : āĻ. āϏāĻžāĻĒā§āĻā§āώ āϏāϰā§āĻŦāύāĻžāĻŽ
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āϞāĻŋāĻā§āĻ: āĻĒāĻĻ āĻĒā§āϰāĻāϰāĻŖ
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ā§§ā§§. āĻĒā§āϰāĻŽāĻĨ āĻā§āϧā§āϰ⧠āϏāĻŽā§āĻĒāĻžāĻĻāĻŋāϤ āϏāĻŦā§āĻāĻĒāϤā§āϰ- āĻĒāϤā§āϰāĻŋāĻāĻž āĻāϤ āϏāĻžāϞ⧠āĻĒā§āϰāĻāĻžāĻļāĻŋāϤ āĻšā§?
      āĻ. ā§§ā§¯ā§§ā§Š
      āĻ. ⧧⧝⧧ā§Ē
      āĻ. ⧧⧝⧧ā§Ģ
      āĻ. ⧧⧝⧧ā§
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āĻāϤā§āϤāϰ : āĻ. ⧧⧝⧧ā§Ē
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⧧⧍. āĻāϰā§āĻŖāĻžāĻāĻžāϞ- āĻļāĻŦā§āĻĻā§āϰ āĻ āϰā§āĻĨ
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      āĻ. āĻĻā§āĻĒāĻžāĻā§āĻāĻž
      āĻ. āĻā§āĻā§āĻāĻāĻŋāĻāĻž
      āĻ. āĻŽāĻžāĻā§āϏāĻžāϞ āϤā§āϰāĻŋ āĻāĻžāϞ
      āĻ. āĻŽāĻžāĻāϧāϰāĻžāϰ āύāĻŋāĻā§āώā§āĻĒāϝā§āĻā§āϝ āĻāĻžāϞ
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āĻāϤā§āϤāϰ : āĻ. āĻŽāĻžāĻā§āϏāĻžāϞ āϤā§āϰāĻŋ āĻāĻžāϞ
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ā§§ā§Š. āĻĒāĻžāĻ ā§āϝ āĻā§āύ āĻāĻŦāĻŋāϤāĻžā§ āĻŦāĻžāϤāĻžāĻŦāĻŋ āύā§āĻŦā§-āϰ āĻāϞā§āϞā§āĻ āĻāĻā§?
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     āĻ. āϤāĻžāĻšāĻžāϰā§āĻ āĻĒā§ā§ āĻŽāύā§
      āĻ. āĻŦāĻžāĻāϞāĻžāĻĻā§āĻļ
      āĻ. āĻāĻāĻāĻŋ āĻĢāĻā§āĻā§āϰāĻžāĻĢ
      āĻ. āĻāĻŦāϰ
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āĻāϤā§āϤāϰ : āĻ. āϤāĻžāĻšāĻžāϰā§āĻ āĻĒā§ā§ āĻŽāύā§
āĻā§āĻāĻž: āĻŦāĻžāϤāĻžāĻŦāĻŋ āύā§āĻŦā§āϰ āĻĢā§āϞ āĻĢā§āĻā§āĻā§ āĻāĻŋ? āĻĢā§āĻā§āĻā§ āĻāĻŋ āĻāĻŽā§āϰ āĻŽā§āĻā§āϞ? (āĻĒā§āϰāĻĨāĻŽ āϏā§āϤāĻŦāĻ)
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ā§§ā§Ē. āĻāĻāĻžāϧāĻŋāĻ āϏā§āĻŦāĻžāϧā§āύ āĻŦāĻžāĻā§āϝāĻā§ āĻāĻāĻāĻŋ āĻŦāĻžāĻā§āϝ⧠āϞāĻŋāĻāϞ⧠āϏā§āĻā§āϞā§āϰ āĻŽāĻžāĻāĻāĻžāύ⧠āĻā§ āĻāĻŋāĻšā§āύ āĻŦā§āϝāĻŦāĻšā§āϤ āĻšā§?
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      āĻ. āĻā§āϞāύ
      āĻ. āĻĄā§āϝāĻžāĻļ
      āĻ. āϏā§āĻŽāĻŋāĻā§āϞāύ
      āĻ. āĻāĻŽāĻž
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āĻāϤā§āϤāϰ : āĻ. āĻāĻŽāĻž
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ā§§ā§Ģ. āĻāĻŽāĻžāĻĻā§āϰ āĻĻāϞ⧠āĻā§āĻš āĻĻāϞāĻĒāϤāĻŋ āύāĻžāĻāĨ¤- āĻāĻžāĻšāĻžāĻĻā§āϰ āĻĻāϞā§?
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      āĻ. āĻāĻŦāĻŋāĻĻā§āϰ
      āĻ. āϤāϰā§āĻŖāĻĻā§āϰ
      āĻ. āϏāĻžāϧāĻāĻĻā§āϰ
      āĻ. āĻŦāĻā§āϤāĻžāĻĻā§āϰ
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āĻāϤā§āϤāϰ : āĻ. āϤāϰā§āĻŖāĻĻā§āϰ
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āĻā§āĻāĻž: āĻāĻŽāĻžāĻĻā§āϰ āĻĻāϞ⧠āĻā§āĻš āĻĻāϞāĻĒāϤāĻŋ āύāĻžāĻ, āĻāĻ āĻāĻŽāϰāĻž āĻļāϤ āĻĻāĻŋāĻ āĻšāĻāϤ⧠āĻļāϤ āĻļāϤ āϤāϰā§āĻŖ āĻŽāĻŋāϞāĻŋā§āĻž āϤāĻžāϰā§āĻŖā§āϝā§āϰ āĻļāϤāĻĻāϞ āĻĢā§āĻāĻžāĻā§āĻž āϤā§āϞāĻŋā§āĻžāĻāĻŋāĨ¤ (āĻāϤā§āϰā§āĻĨ āĻ āύā§āĻā§āĻā§āĻĻ)
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ā§§ā§Ŧ. The situation has come to a head- āĻāϰ āĻ āϰā§āĻĨ-
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      āĻ. āĻĒāϰāĻŋāϏā§āĻĨāĻŋāϤāĻŋāϰ āĻāύā§āύāϤāĻŋ āĻāĻā§āĻā§
      āĻ. āĻĒāϰāĻŋāϏā§āĻĨāĻŋāϤāĻŋ āϏāĻŦāĻā§ā§ā§ āĻāĻžāϞ⧠āĻ āĻŦāϏā§āĻĨāĻžā§ āĻāϏ⧠āĻĻāĻžāĻā§āĻŋā§ā§āĻā§
      āĻ. āĻĒāϰāĻŋāϏā§āĻĨāĻŋāϤāĻŋāϰ āĻ āĻŦāύāϤāĻŋ āĻāĻā§āĻā§
      āĻ. āĻĒāϰāĻŋāϏā§āĻĨāĻŋāϤāĻŋ āĻāϰāĻŽ āĻ āĻŦāϏā§āĻĨāĻžā§ āĻĒā§ā§āĻāĻā§āĻā§
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āĻāϤā§āϤāϰ : āĻ. āĻĒāϰāĻŋāϏā§āĻĨāĻŋāϤāĻŋ āĻāϰāĻŽ āĻ āĻŦāϏā§āĻĨāĻžā§ āĻĒā§ā§āĻāĻā§āĻā§
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ā§§ā§. āĻŽāĻŖā§āĻĄāĻ- āĻļāĻŦā§āĻĻā§āϰ āĻ āϰā§āĻĨ-
      āĻ. āĻŽā§āĻŖā§āĻĄāϧāĻžāϰā§
      āĻ. āĻā§āĻ
      āĻ. āĻā§ā§ā§
      āĻ. āĻŽāĻŋāώā§āĻāĻŋāĻāĻžāϤā§ā§ āĻĻā§āϰāĻŦā§āϝ
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āĻāϤā§āϤāϰ : āĻ. āĻā§āĻ
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ā§§ā§Ž. āύāĻŋāĻā§āϰ āĻā§āύāĻāĻŋ āĻļāĻžāĻŽāϏā§āϰ āϰāĻžāĻšāĻŽāĻžāύā§āϰ āĻāĻžāĻŦā§āϝāĻā§āϰāύā§āĻĨ āύā§?
      āĻ. āĻĻā§āĻāϏāĻŽā§ā§āϰ āĻŽā§āĻā§āĻŽā§āĻāĻŋ
      āĻ. āĻāĻĻā§āĻāĻ āĻāĻā§āϰ āĻĒāĻŋāĻ ā§ āĻāϞā§āĻā§ āϏā§āĻŦāĻĻā§āĻļ
      āĻ. āύāĻŋāĻ āĻŦāĻžāϏāĻā§āĻŽā§
      āĻ. āĻāĻāĻ āϏāύā§āϧā§āϝāĻžā§ āĻŦāϏāύā§āϤ
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āĻāϤā§āϤāϰ : āĻ. āĻāĻāĻ āϏāύā§āϧā§āϝāĻžā§ āĻŦāϏāύā§āϤ
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⧧⧝. āĻā§āύ āĻļāĻŦā§āĻĻāĻāĻŋ āĻāĻĒāϏāϰā§āĻāϝā§āĻā§ āĻāĻ āĻŋāϤ?
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      āĻ. āĻ āĻŦāϰā§āĻŖā§āϝ
      āĻ. āϤāϰā§āĻŖ
      āĻ. āĻĒāϰā§āĻā§āώāĻž
      āĻ. āĻāϞā§āώ
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āĻāϤā§āϤāϰ : āĻ. āĻ āĻŦāϰā§āĻŖā§āϝ
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āϞāĻŋāĻā§āĻ: āĻāĻĒāϏāϰā§āĻ
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⧍ā§Ļ. āϏāĻžāĻšāĻāϰā§āϝ- āĻļāĻŦā§āĻĻāĻāĻŋ āĻāĻ āĻŋāϤ āĻšā§ā§āĻā§
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     āĻ. āĻĒā§āϰāϤā§āϝā§āϝā§āĻā§
      āĻ. āϧāĻžāϤā§āϝā§āĻā§
      āĻ. āϏāύā§āϧāĻŋāϝā§āĻā§
      āĻ. āϏāĻŽāĻžāϏāϝā§āĻā§
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āĻāϤā§āϤāϰ : āĻ. āĻĒā§āϰāϤā§āϝā§āϝā§āĻā§
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āϞāĻŋāĻā§āĻ: āĻĒā§āϰāĻā§āϤāĻŋ-āĻĒā§āϰāϤā§āϝā§
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⧍⧧. āύāĻŋāϰāĻžāĻāĻžāϰ- āĻļāĻŦā§āĻĻā§āϰ āϏāύā§āϧāĻŋāĻŦāĻŋāĻā§āĻā§āĻĻ āĻā§āĻžāύāĻāĻŋ?
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      āĻ. āύāĻŋ+āĻāĻāĻžāϰ
      āĻ. āύāĻŋāĻ+āĻāĻāĻžāϰ
      āĻ. āύāĻŋāϰ+āĻāĻāĻžāϰ
      āĻ. āύāĻŋāϰāĻ+āĻāĻāĻžāϰ
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āĻāϤā§āϤāϰ : āĻ. āύāĻŋāĻ+āĻāĻāĻžāϰ
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⧍⧍. āĻāϰā§āĻŦā§āϰ- āĻļāĻŦā§āĻĻā§āϰ āĻ āϰā§āĻĨ
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      āĻ. āϰāĻžāĻā§āώāϏ
      āĻ. āĻāύā§āϧāĻĻā§āϰāĻŦā§āϝāĻŦāĻŋāĻļā§āώ
      āĻ. āϰāĻžāϏāĻžā§āύāĻŋāĻ āĻĒāĻĻāĻžāϰā§āĻĨ
      āĻ. āĻāϰāĻŖā§ā§
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āĻāϤā§āϤāϰ : āĻ. āϰāĻžāĻā§āώāϏ
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ā§¨ā§Š. āϤā§āώāĻžāϰāϧāĻŦāϞ- āĻā§āύ āϏāĻŽāĻžāϏā§āϰ āĻāĻĻāĻžāĻšāϰāĻŖ?
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      āĻ. āϏāĻžāϧāĻžāϰāĻŖ āĻāϰā§āĻŽāϧāĻžāϰā§
      āĻ. āĻāĻĒāĻŽāĻžāύ āĻāϰā§āĻŽāϧāĻžāϰā§
      āĻ. āĻāĻĒāĻŽāĻŋāϤ āĻāϰā§āĻŽāϧāĻžāϰā§
      āĻ. āĻŽāϧā§āϝāĻĒāĻĻāϞā§āĻĒā§ āĻāϰā§āĻŽāϧāĻžāϰā§
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āĻāϤā§āϤāϰ : āĻ. āĻāĻĒāĻŽāĻžāύ āĻāϰā§āĻŽāϧāĻžāϰā§
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āϞāĻŋāĻā§āĻ: āϏāĻŽāĻžāϏ
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⧍ā§Ē. āĻŖ-āϤā§āĻŦ āĻŦāĻŋāϧāĻŋ āĻ āύā§āϏāĻžāϰ⧠āĻā§āύ āĻā§āĻā§āĻ āĻ āĻļā§āĻĻā§āϧ āĻŦāĻžāύāĻžāύā§āϰ āĻĻā§āώā§āĻāĻžāύā§āϤ?
      āĻ. āϧāϰāύ, āĻŦāϰāĻŖ
      āĻ. āĻŦāϰā§āύāύāĻž, āĻĒā§āϰā§āύā§
      āĻ. āύā§āϤā§āϰāĻā§āύāĻž, āĻĒāϰāĻāύāĻž
      āĻ. āϰā§āĻĒāĻžā§āĻŖ, āĻĒā§āϰāĻŖā§āύ
āĻā§āĻāĻž: āĻĻā§āĻŦāĻŋāϤā§ā§ āĻā§āĻā§āĻā§āϰ āĻļā§āĻĻā§āϧāϰā§āĻĒ- āĻŦāϰā§āĻŖāύāĻž, āĻĒā§āϰāύā§
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āĻāϤā§āϤāϰ : āĻ. āĻŦāϰā§āύāύāĻž, āĻĒā§āϰā§āύā§
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āϞāĻŋāĻā§āĻ: āĻŖāϤā§āĻŦ āĻ āώāϤā§āĻŦ āĻŦāĻŋāϧāĻžāύ
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⧍ā§Ģ. āϏāĻŽāĻāĻŋāĻŦā§āϝāĻžāĻšāĻžāϰ- āĻļāĻŦā§āĻĻāĻāĻŋāϤ⧠āĻŽā§āĻ āĻā§āĻāĻŋ āĻāĻĒāϏāϰā§āĻ āĻāĻā§?
      āĻ. ⧍
      āĻ. ā§Š
      āĻ. ā§§
      āĻ. ā§Ē
āĻā§āĻāĻž: āϏāĻŽ+āĻ āĻāĻŋ+āĻŦāĻŋ+āĻšāĻžāϰ (āϏāĻŽ, āĻ āĻāĻŋ, āĻŦāĻŋ- ā§ŠāĻāĻŋ āĻāĻĒāϏāϰā§āĻ)
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āĻāϤā§āϤāϰ : āĻ. ā§Š
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āϞāĻŋāĻā§āĻ: āĻāĻĒāϏāϰā§āĻ
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⧍ā§Ŧ. āĻāĻžāĻŦā§āϝ āĻāĻāϤ⧠āϝāĻžāϰ āύāĻžāĻŽ āĻāύāύā§āĻĻ āϤāĻžāϰā§āĻ āύāĻžāĻŽ āĻŦā§āĻĻāύāĻžāĨ¤â āĻŦāĻžāĻā§āϝāĻāĻŋ āĻāĻā§ āϝ⧠āϰāĻāύāĻžā§-
      āĻ. āĻŦāĻŋāϞāĻžāϏā§
      āĻ. āĻšā§āĻŽāύā§āϤā§
      āĻ. āϏāĻžāĻšāĻŋāϤā§āϝ⧠āĻā§āϞāĻž
      āĻ. āϝā§āĻŦāύā§āϰ āĻāĻžāύ
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āĻāϤā§āϤāϰ :  āĻ. āϏāĻžāĻšāĻŋāϤā§āϝ⧠āĻā§āϞāĻž
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⧍ā§. āĻāĻžāϰ āĻŽā§āĻā§ā§āĻžāĻ āĻāϰāĻž āĻĻāĻžāĻāϤ āĻŦāĻžāϞāĻŦ-āĻāϰ āĻāϞā§ā§ āĻāĻāĻāĻ āĻāϰā§?
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      āĻ. āĻŽā§āĻĻāĻžāĻŦā§āĻŦā§āϰā§āϰ
      āĻ. āĻŽāĻāϏā§āĻĻā§āϰ
      āĻ. āĻāĻāύā§āϏā§āϰ
      āĻ. āϰāĻžāĻšāĻžāϤā§āϰ
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āĻāϤā§āϤāϰ : āĻ. āĻŽā§āĻĻāĻžāĻŦā§āĻŦā§āϰā§āϰ
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āĻā§āĻāĻž: -āĻāϰāĻžāĻ āϤ⧠āϏāĻŦāĻāĻŋāĻā§āϰ āĻŽā§āϞā§, āĻŽā§āĻĻāĻžāĻŦā§āĻŦā§āϰ āĻŦāϞā§āĨ¤ āĻāϞāĻā§āĻ āĻŦāĻžāϞāĻŦ-āĻāϰ āĻāϞā§ā§ āϤāĻžāϰ āϏāϝāϤā§āύ⧠āĻŽā§āĻā§ā§āĻžāĻ āĻāϰāĻž āĻĻāĻžāĻāϤ āĻāĻāĻāĻ āĻāϰā§āĨ¤ āϤāĻžāĻĻā§āϰ āύā§āĻāϤāĻž āĻšā§āύāϤāĻž āĻā§āĻā§āĻžāĻŽāĻŋāϰ āĻāύā§āϝā§āĻ āϤ⧠āĻĻā§āĻļāĻāĻž āĻāĻžāĻ āĻšāϞā§āĨ¤
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ā§¨ā§Ž. Lyric āĻļāĻŦā§āĻĻā§āϰ āĻŦāĻžāĻāϞāĻž āĻĒāϰāĻŋāĻāĻžāώāĻž=
      āĻ. āĻāĻžāύ
      āĻ. āϏā§āϰ
      āĻ. āĻāĻĨāĻž
      āĻ. āĻā§āϤāĻŋāĻāĻŦāĻŋāϤāĻž
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āĻāϤā§āϤāϰ : āĻ. āĻā§āϤāĻŋāĻāĻŦāĻŋāϤāĻž
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⧍⧝. Vertical āĻļāĻŦā§āĻĻā§āϰ āĻĒāϰāĻŋāĻāĻžāώāĻž-
      āĻ. āĻ āύā§āĻā§āĻŽāĻŋāĻ
      āĻ. āĻāĻā§āĻāϤāĻž
      āĻ. āĻāϞā§āϞāĻŽā§āĻŦ
      āĻ. āĻĻā§ā§āĻžāϞ
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āĻāϤā§āϤāϰ : āĻ. āĻāϞā§āϞāĻŽā§āĻŦ
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ā§Šā§Ļ. āĻĒāĻŋāĻĒāĻžāϏāĻŋāϤ āĻļāĻŦā§āĻĻā§āϰ āĻŦāĻŋāĻļā§āώā§āϝāϰā§āĻĒ-
      āĻ. āĻĒāĻŋāĻĒāĻžāϏā§
      āĻ. āĻĒāĻŋāĻĒāĻžāϏā§
      āĻ. āĻĒāĻŋāĻĒāĻžāϏāĻž
      āĻ. āĻĒāĻŋā§āĻžāϏā§
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āĻāϤā§āϤāϰ : āĻ. āĻĒāĻŋāĻĒāĻžāϏāĻž
āĻŦāĻžāĻāϞāĻž ā§§ā§ āύāĻ āĻāϰ āĻāϤā§āϤāϰ āĻā§āύāĻāĻŋ ?
āĻŦāĻžāĻāϞāĻž ā§§ā§ āύāĻ āĻāϰ āĻāϤā§āϤāϰ āĻā§āύāĻāĻŋ ?
physics
5 & 11 no ans are incorrect.....5(āĻ)  & 11 (āĻ)
physics
vaia sound diffraction show kre bt plarisation shoe kre na........source shajahan tapan sir er boi.......
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boltsman's constant er akok J/K (source phy 1,tofajjol sir)
r hidrogen poromanur shoktistor er ans to 10.2ev hoar kotha,as energy absorvation E=E2-E1?
ENGLISH 25 No. er ans Wrong!!
25.
(A) The jury are arguing among themselves.
(B) The jury is arguing among themselves.
(C) The jury has argued among themselves.
(D) The jury has been arguing among themselves.
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Answer: (B) The jury is arguing among themselves.
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Jury er por is boshe.MS word a dilei correct kore dibe.
Physics
Ans. A
āĻāĻŖāĻŋāϤ āĻāϰ ⧍⧠āύāĻ āĻĒā§āϰāĻļā§āύā§āϰ āĻāϤā§āϤāϰ
āĻāĻŖāĻŋāϤ āĻāϰ ⧍⧠āύāĻ āĻĒā§āϰāĻļā§āύā§āϰ āĻāϤā§āϤāϰ āĻā§āϞ āϰā§ā§āĻā§ āĨ¤ āĻĒā§āϰāĻļā§āύ⧠āĻāĻāĻāĻŋ āĻŽā§āĻāĻž āĻĒāϰāĻžāϰ āĻāĻĨāĻž āĻŦāϞāĻž āĻšā§ā§āĻā§ āĨ¤ āĻāĻŋāύā§āϤ⧠āĻāĻ āϏāĻŽāĻžāϧāĻžāύ⧠āĻāĻ āĻā§āϰāĻž āĻŽā§āĻāĻž āĻšāĻŋāϏāĻžāĻŦ āĻāϰāĻž āĻšā§ā§āĻā§ āĨ¤
chemistry
chemistry 17 number question er ans b hobe.
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12.āĻ) 10.2eV 13.6-3.4=10.2 eV
12.āĻ) 10.2eV
13.6-3.4=10.2 eV
20. 50/20=2.5
20. 50/20=2.5
biology
bio 8 num ki thik ace????
13. Iâd like ------
13. Iâd like ------ information, please.
(A) an
(B) some
(C) few
(D) piece
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āĻāĻāĻž āϤ⧠āĻŽāύ⧠āĻšā§ Few āĻšā§āĻžāϰ āĻāĻĨāĻžāĨ¤ āĻāĻŽāĻžāϰ āĻāĻžāύāĻž āĻŽāϤ⧠Information āĻāϰ āĻāĻā§ Few āĻŦāϏā§āĨ¤
ā§§ āύāĻ āĻĒā§āϰāĻļā§āύ (physics)
ā§§ āύāĻ āĻĒā§āϰāĻļā§āύā§āϰ āĻŦā§āϝāĻžāĻā§āϝāĻžā§ āĻā§āϞ āϰā§ā§āĻ⧠⧎
MATH āĻāϰ PHYSICS āϏāĻ āĻŋāĻ āĻāϤā§āϤāϰā§āϰ
MATH āĻāϰ PHYSICS āϏāĻ āĻŋāĻ āĻāϤā§āϤāϰā§āϰ āĻĒāĻžāĻļāĻžāĻĒāĻžāĻļāĻŋ āϏāĻŽāĻžāϧāĻžāύ āĻĻāĻŋā§ā§ āĻĻā§āĻā§āĻžāϤ⧠āϧāύā§āϝāĻŦāĻžāĻĻ!!Â
chemistry
⧍⧝ āύāĻ āĻĒā§āϰāĻļā§āύā§āϰ āĻāϤā§āϤāϰ C J/k
āĻĢāĻŋāĻāĻŋāĻā§āϏ
tnx a lot
āύāϤā§āύ āĻāĻŽā§āύā§āĻ āϝā§āĻā§āϤ āĻāϰā§āύ